What is the range of degeneracy pressure? In what way is degeneracy pressure related to the separation of fermions? Is there any influence at ranges like a meter and beyond. I expect the influence at those ranges to be immeasurably small. I just wonder if theory predicts an effect or if there is truly zero degeneracy pressure at long range.
I'm under the impression that if you have a block of iron that is a solid cubic meter, then each and every electron in that block will be found in a slightly different quantum state at any given moment. This being due to the PEP.
 A: If you are considering ideal fermion degeneracy pressure, then the answer is simple.
If the fermions are non-relativistic, the degeneracy pressure scales as $n^{5/3}$, where $n$ is the fermion number density.  Thus for a fixed number of fermions the pressure decreases more rapidly than the volume increases.
If the fermions are not completely degenerate then the "contribution of degeneracy" to the pressure (it cannot really be separated as such) is lower, and of course the fermions do become less degenerate as you reduce their density.
So in principle there is always some degeneracy pressure, but it would quickly become negligible.
However, this discussion is about free fermions. Your question appears to be about atoms. I have never come across degeneracy pressure being explicitly discussed in atomic material, so am unable to properly address your question I think. The force between atoms is often represented by models such as the Lennard-Jones or Morse potentials. These are repulsive at short range, presumably as a result of overlapping electronic wavefunctions, but are small and attractive at long range. Of course all this means is that any repulsion due to the PEP at long range is overwhelmed by other (attractive) effects.
