As the proton tunneling frequency is a measurable quantity, so the associated vibrational relaxation time can be determined. Therefore, the proton tunneling rate would be a kinetics relative value depending on temperature, solvent, hydrogen bonding distance and topology. As Benabbas et al. described in the following publication, the proton tunneling takes place on sub nano-second timescale at room temperature for typical H-bonding distance. Hope I well understood your query.

http://dx.doi.org/10.1063/1.4913591

The question doesn't make sense, because the barrier doesn't exist for the proton-it's a (semi)classical artifact. It doesn't forbid propagation, just changes how it occurs. The barrier just means that the proton isn't a free particle. The amplitude for the proton's propagation ``across the barrier'' has a particular dependence on the coupling constant (or Planck's constant), that can be determined.  Nothing mysterious. it's no more or less sudden than what it takes for the proton to propagate from any one point to any other point; simply that the potential, actually felt by the proton, isn't the classical potential-it doesn't have any barrier, but is more ``complicated'' than the classical potential. 

The confusion arises because, if one were to study the *classical* limit of the process, one would find it, indeed, impossible to describe-but this, simply, is due to the fact that a classical limit doesn't exist. In the classical limit the barrier is a ``classically forbidden region''.  This simply means that a classical probe will never find the proton in this region. 

I`d say that nothing can occure suddenly: each velocity has an asymptotically barrier.

Though wave function has a time dependence, if you plot probability density with respect to the space coordinate, it will not have any time dependence. In usual text books one discusses two separate cases which are E < V0 and E > V0, where E is the energy of the point particle under consideration and V0 is the height of the barrier. In the first case a classical particle has "no chance" of existing at the other side of a barrier, whereas for a quantum particle there exists a non-zero probability for it to be at the other side. This is a result of describing this situation in terms of the Schrodinger equation.

A more general way would be to use time-energy uncertainty principle. The subtle point to remember is that to start with the Schrodinger equation has been derived using uncertainty relations. In this case one can use time-energy uncertainty relation ΔE Δt ~ ħ to estimate the degree of "non-sudden" behavior.

A discussion can be found at: http://www.physicsoftheuniverse.com/topics_quantum_uncertainty.html

Wrong-the time dependence of the wavefunction-relevant for the problem at hand, in particular-obviously leads to the time dependence of the probability density! Of course it's possible to write the Schrödinger equation as an eigenvalue equation-but one must then perform the sum over the eigenvalues. If the initial condition is an eigenstate of the (full) Hamiltonian then the wavefunction varies by a phase, that cancels when taking the absolute value-but all initial conditions are not eigenstates of the full Hamiltonian. And the eigenstates of the full Hamiltonian are not eigenstates of the kinetic energy, i.e. the Hamiltonian of the free particle, since position and momentum don't commute. 

Stam, as usual the response to your criticism is that there was nothing wrong as such when a single wave function with a definite eigenvalue is taken at a time. If you take two different wave functions ψ1 and ψ2  for two different eigenvalues  which are E1 and E2, then obviously in |ψ1 + ψ2|2  time dependent phase will not cancel, and one gets an interference term which is time dependent. But do you have something more to say? It was unclear to me.

It is not just possible to write Schrodinger equation as an eigenvalue equation as you say, it is the most common practice.