Only in the smallest scalar Universe, the smallest relativistic model of 3 points, is there a case of partitioning where we can say it is finite. What is curious here is that we can have this with virtual particles that are present but perhaps unmeasurable, plus the minimum relativistic 3 particles. Add to this that the number of fermions is going to be a triad (odd number of finite point particles, and we can be certain that even this system is not at rest. That necessarily implies the fermions are also in a periodic system of transport. While even the simplest scenario is complicated. it is a great academic example to start from.

The fact that the volume operator has discrete spectrum, doesn't imply that its excitations can be, necessarily, identified with fermions (or bosons) at all-this property of the spectrum doesn't determine their statistics. Nor does this property of the spectrum of this operator have, as such, anything to do with the density of the excitations of any other operator. The question attempts to relate two issues that don't have anything to do with each other. 

That's why attempting to postulate any such relation, absent any further information, doesn't make sense.

(Of course, for fermions-assuming they can be defined-to condense, an additional, attractive, interaction is necessary. The caveat is that to speak meaningfully of fermions, or bosons, at all, a spin-statistics relation is required, which isn't, always, available: the spin-statistics theorem has assumptions: http://ejde.math.txstate.edu/conf-proc/04/w1/wightman.pdf) 

It's, of course, a triviality that, in the presence of a cutoff, the number and the density of any quantity, will be finite. The non-trivial statement is whether, for any given quantity,  a limit exists, that's independent of how the cutoff is defined and  how it is removed. 

The reason the volume operator has a discrete spectrum is because  the group, namely SU(2), from whose generators it's constructed, is compact. So any statement that this fact is relevant for the description of quantum effects requires further elaboration. If it would take values in the algebra, for instance, some more work would have been required: the algebra is a non-compact manifold. 

That all operators, whose domain is a compact manifold, have discrete spectrum, doesn't have anything to do with any quantization issue-but is a (mathematical) statement, that's independent of Planck's constant. 

Conversely, that, under certain assumptions, there do exist operators, with continuum spectrum, whose domain is a non-compact manifold, that do describe quantum effects, i.e. that depend non-trivially on Planck's constant, is, also, true. 

So a discrete spectrum doesn't, necessarily, imply the relevance of quantum effects, nor does a continuous spectrum imply their irrelevance. 

A non-trivial problem is to compute the degeneracy of the corresponding spectra-especially when the spectrum is continuous-and show that it is a well-defined quantity, i.e. doesn't depend on the cutoff procedure used to define it in the first place-and can be used for well-defined calculations. For the degeneracy is a completely independent property of the spectrum.

Curiously, these facts aren't  as well-known as might be expected.