What balances the electron degeneracy pressure in a solid crystal? In a solid crystal, for example, copper, if we treat the electrons as a free electron gas we can obtain that the pressure exerted by the gas at absolute zero is about $10^5$ atm or $10$ GPa. What balances the electron degeneracy pressure in a solid?
Furthermore, if we apply external pressure greater than $10$ GPa, sometimes it will only lead to a phase transition rather than destroying the crystal. What is the force that balances the external pressure? Is it still the electron degeneracy pressure that balances the external pressure?
 A: The obvious answer here is that the degeneracy pressure is balanced by electrostatic attraction forces due to the positively charged nuclei. This is same idea for any bound fermion system, the degeneracy pressure that makes fermions stick together always is balanced by whatever is attracting them in the first place. 
For protons and neutrons this is the (residual) strong force. 
For electrons it is the EM force. 
For neutron stars it is gravity.
A: "...What balances the electron degeneracy pressure in a solid?..."
The problem with this statement is that it assumes that the "electron gas" is exactly analogous to, say, Helium in a balloon. We need to (in my opinion) find slightly different ways of thinking of Pressure of a solid, as against pressure of a fluid.
Helium in a balloon is a fluid. What that means is that if we say that the Pressure of a Helium balloon in equilibrium with its surroundings is $P$, then, we can conclude that the pressure in the "atmosphere" surrounding the balloon must also be $P$. The fact that we are dealing with something that is able to flow implies that unless balanced by an equal and opposite pressure, the fluid will "spread" till its pressure equals that of its surroundings.
Let us now compare this with the situation you describe. That is, a lattice (a rigid solid, definitely cannot flow); On this background, a "sea" of nearly free electrons. Obviously, this whole setup is in equilibrium as we see it, notwithstanding any Pressure it has. 
Note that we may bring in atmospheric pressure into this, but I think that is misleading. I am pretty sure a piece of metal (which is an example of a crystalline solid with a degenerate electron gas) won't disintegrate if it is placed in a vacuum.
So how should we think of pressure in a solid? Strictly speaking, using the thermodynamic relation $dE = -PdV$ ; ie., Pressure is the rise in Internal energy of the solid per unit volume of compression. If we compress/stretch the solid, then energy rises/falls by $P$ per unit volume. Compare this with the physical implication of pressure of a fluid above. Solids deform via intentionality. Fluids, in contrast, deform spontaneously, just because they can flow. 
Seen in this way, it is obvious why the electron-gas should have a "Pressure". If we compress the system, we are changing the boundary conditions for the electronic states, (occupied by the electrons forming the gas) and hence its internal energy. In short, Pauli exclusion principle ; the energies of the levels change, but their occupation numbers can't (at absolute $0$), so the total energy changes.
Also, note that this is not the only source of "Pressure" in this system. Obviously, the rigid lattice has its own associated pressure as well. But none of this means that this needs to be balanced by external pressures in order for it to be in equilibrium.
Addendum, on nomenclature: The exchange of comments below also made me realize, that the nomenclature of electron gas or fluid may be confusing. An electron fluid is so called because it is fluid with respect to transport of energy, charge, spin, etc (whatever the electrons can transport). But an electron fluid is definitely not to be thought of as fluid with respect to mechanical deformations. If I had to say what the other answer is saying (from what I understand), is that the reason the electron fluid does not require an external pressure to hold it together is the attraction of the ionic lattice. Frankly, I find this a little confusing, because the fluidity of a so called electronic fluid is very much an emergent phenomenon specific to this special situation of electrons on a lattice. What I mean is, normally, one starts with a tight-binding model, then takes the long wavelength (low energy) limit ; then, a fluid emerges. I don't know how to make sense of this outside of this specific context.
So, the safest thing to say, it seems, is that the electron fluid is not fluid in a mechanical sense because it is tied to a mechanically rigid elastic medium (the lattice). How exactly it is tied is not at all a factor in the behaviour of the electron fluid. Remember, that the tying of free electrons to the lattice happens on the length scale of the lattice. The fluid emerges at a much larger lengthscale, by which point, the exact nature of the underlying rigid lattice is irrelevant. All that matters is the fluid is tied to some elastic solid, so its mechanical deformations are constrained to this unspecified background solid. That is all that is required to explain why the degeneracy pressure of the electron fluid is not to be thought of as something that needs to be counterbalanced by an equal and opposite pressure to prevent the electron fluid from flowing out.
Addendum (2): A comment points out below, the "nearly free electrons" is a poor approximation : It is more subtle that that ; In most normal metals, the nearly free electron model IS a very good model of the macroscopic properties of what we see (with renormalized parameters, of course). That is why the Boltzmann equation works so well in calculating transport coefficients in normal electron fluids. (leading to the Fermi - liquid theory). Of course, this fails sometimes, in quantum critical systems, which is when we have various kinds of Non-Fermi liquid behaviour. These phases are marked by the absence of quasiparticle excitations of any kind. Hence, it is objectively true in these situations that the model of nearly free electrons is a poor one to explain the macroscopic properties we see. But normally (in the aptly named normal fermi liquid phase) , the nearly free-electron model is an astonishingly good one (again, to explain macroscopic properties of the fluid)
A: tl;dr (rigidity vs positivity of background): 
Several comments on this post have pointed out the background of positive ions forming the lattice. Few points about this:
$(1)$ At the scale at which we observe an electron liquid, this appears as a uniform positive background. In an isolated crystal (which is electrically neutral) this constrains the density of the electron fluid to be the density of the background. But this is a trivial fact that cannot really be used to explain anything. $(3)$
$(2)$ But first, let us reiterate that definitely, invoking "positively charged nuclei" to explain why the electron fluid does not flow out under the influence of its own pressure is a gross simplification (that may, as I have mentioned several times on this post, be very misleading). There are no discrete centers of positive charge (nuclei) that the electron fluid sees.
$(3)$ But might we at least use the fact of the uniform positive background to explain the fact that the electronic fluid does not flow out on account of its own pressure? Answer is again, no. For one, A Uniformly Charged Background Exerts no Forces (Obviously). 
If that is not convincing, consider that a crystal that is Not Isolated. We are usually allowed to tune the chemical potential of the crystal as well. So we can create a situation where the (uniform background + electron fluid) system is definitely NOT neutral ; still, nothing flows out because of "excess pressure"
$(4)$ Why not? If we think in terms of what the other answer (and other comments) are suggesting, For each configuration of electrons, we would have to think of a source of equal and opposite pressure. The reason this seems tedious is that it is wrong. It seeks to explain an inherently quantum phenomenon in terms a classical toy model.
$(5)$ A normal metal (system in OP's question) (normal $\implies$ quasi-particles exist) is very much a quantum fluid. It has a degeneracy pressure, relating to the fact that (as I mentioned in my other answer):  $(I)$  the quasi-particle excitation energies are related to the dimensions of the crystal and $(II)$ Pauli Exclusion principle
This does imply, as I referred to in my other answer, that we have an operational definition of $Pressure.$ However it does NOT imply that we need to be worried about the electron fluid spreading out unless counterbalanced by an equal and opposite pressure (to reiterate). 
$(6)$ So what is the key operative reason why the electron fluid does not flow out? As we saw above, it is that the electron fluid is an emergent phenomenon on a Rigid background. 
This seems to really be the theme of the whole discussion that has gone before. Rigidity of background vs Positivity of background. 
I hope I have demonstrated that correct reason is the former. 
