As far as I can see, no. The reason is that any two particles could come very close together, thus acquiring a very large negative energy. This energy, if carried over as positive kinetic energy to the third particle could always be made so large as to allow escape. There is a difficuly, however: in some initial configurations, it may be either very difficult or plain impossible to reach such tightly bound two-body configurations, in which the third body escapes. But I do not think one can give a value of the energy, below which all particles remain bound. In fact one has an opposite result, linked to Poincaré's recurrence theorem: if any three body configuration could arise from one particle coming from infinity in the wo-particle system, then with probability one this configuration will also dissociate in the future.

As far as I can see, no. The reason is that any two particles could come very close together, thus acquiring a very large negative energy. This energy, if carried over as positive kinetic energy to the third particle could always be made so large as to allow escape. There is a difficuly, however: in some initial configurations, it may be either very difficult or plain impossible to reach such tightly bound two-body configurations, in which the third body escapes. But I do not think one can give a value of the energy, below which all particles remain bound. In fact one has an opposite result, linked to Poincaré's recurrence theorem: if any three body configuration could arise from one particle coming from infinity in the wo-particle system, then with probability one this configuration will also dissociate in the future.

Yes.  The escape velocity is the minimum speed one body must achieve to leave the orbit of the other two.  As you posed the problem with all 3 bodies having the same mass, the escape velocity is the equation for the two body problem, where 2 of the bodies are combined, and use the easier 2 body escape velocity for the lighter body.

That said, while it's a righteous answer to your initial problem statement, it may not be what you intended.  Regarding your qualifying paragraph, I have the below to add.

It is known that any 3 body orbital solution always ends up with one of the three bodies being ejected, at a speed often above the escape velocity, by a multiple, or even more an order of magnitude.  Sooner, not later.  A lighter body has a greater chance of being ejected than the heaviest. 

To keep all the bodies in orbit, with none escaping is possible, but one must set the initial positions and velocities to be stable, with no 'interference.'  With heavenly bodies this 'setting' is not possible. In a computer simulation, yes, it's possible.

One such known stable orbit is two of the bodies orbit each other 'closely', while the third body is far away, enough to see the inner two bodies as a single center of gravity (which it is).  I believe the orbit distance of the third body must be greater than one order of magnitude, if not two, from the inner two bodies.

I hazily recall reading a proof of calculating the outer orbit as a minimum distance, given all 3 masses, and the inner orbit distance.

There are known 3 star systems with this configuration, but they have different masses.

To complete my answer, there is no minimum energy for the orbit of 3 masses that are all 'close' to each other.  It's inherently unstable, sooner, not later, meaning within hundreds of orbits, if not dozens, or quicker, one body will be ejected.  Unless, the bodies are started adequately far away from each other.  The left over two body may be stable, but it's not required.  That is, the ejected body's mass may destabilize the inner two body's orbit upon leaving, the two inner bodies will collide.

Look up "rogue stars", leaving our galaxy, and the proposed mechanism that ejects them at such high velocity, to get them out of the galactic plane. More than a dozen have been spotted now, compared to ten years ago (zero then).

I got this information from my years in college, from a professor who said this.  And proved it on the black board.  And from reading NASA.gov updates.

Peter I suppose if you apply Gauss' theorem to the class of cases where the third particle at a given time lies outside a sphere occupied by the other two, this argument holds? 

In summary, there is no easy criterion to know whether a three-particle system will dissociate or not. In general, most of them do. If you want to make a system which will take at least a avery long time in dissociating, you should take, as correctly stated by Peter, two rather tightly bound bodies, and a third one with a near circular orbit around both and at a large distance. Note that this description is by no means in terms of energy. Indeed, it is is not so important that the particle pair be very tightly bound, it is more important that the third particle should have a high angular momentum, and hence even be rather loosely bound. 

I'm thinking more perturbation theory, placing the third body far enough away, that the orbits of the inner two bodies is seen as a 'vibrating' force for the third body, causing the third body's orbit to deviate from circular.  If the deviation grows, diverges, then the third body is orbiting too close.  If the deviation damps out, then too far.  There should be an inflection point between these two states, where one could say, that is the minimum orbit distance for stability.

Note, there are three cases.  The orbital planes are coplanar.  The inner orbital plane is at right angles to the third body orbital plane.  And the angles in between.  Solutions would be different for each of these cases.  Well, one might be easier to solve, the coplanar case.

This inner/outer situation is not the only 'stable' 3 body situation.  There are others.  Some are stable, like balancing a pin on it's point.  Take the equal lateral triangle case, if the bodies are far enough apart, and only gravity the force in effect, the orbit could last very long.  Or the case, more well known, of the Sun, Earth and Moon, but those are not equal mass.

If running computer simulations do not forget to include the radius of the body.  As they get closer, gravity is no longer a point source, and collisions upset orbits.  If the mass is high enough, then the inner two bodies will collide, like two black holes do.

You may be interested in this article.

http://news.sciencemag.org/physics/2013/03/physicists-discover-whopping-13-new-solutions-three-body-problem

I'm not really sure if it's real science though, playing about on a computer discovering new orbits, there may well be an infinite number of distinct types of closed orbits.

The equilateral triangle orbit is actually unstable whenever the bodies do not have very different masses. In the case of the Sun, Jupiter and the Trojan asteroids, the orbit is stable, because of the large mass ration between the Sun and Jupiter. The ineer/outer case would become ever more stable as the outer orbit increases. However, and this is important, I am not at all sure that all configurations are *for ever* stable, even in this perturbation theoretic case. It might last a very very long time, say exponentially in some large parameter, but there is always a risk (particularly if you allow the orbits not to be planar) that eventually the system dissociates.

Here is a link to a downloadable copy of a very readable textbook. The question I asked is answered in Chapter 8.

http://www.observatorio.unal.edu.co/paginas/docentes/Libro3-Karttunen.pdf

I think that if the initial position of the three particles and the interaction laws are given, then we can define the a minimum of system kinetic energy below which none of the three particles will escape. Otherwise, if the initial conditions are of any kind than we can not.   

Bejo: I disagree. For example, if two of the particles are nearby and the other distant, then if the distant particle has a near circular orbit, then the three  particles will remain bound for an extremely long time (though perhaps not forever). If on the other hand the two particles turn around each other and the third falls, with essentially zero angular momentum, on the bound particles, the system is extremely lilkely to dissociate. The crucial point is, that the way in which the kinetic energy is distributed, as well as other issues involving angular momentum, are important. Given an initial position, things will depend on other things than the kinetic energy