I've wondered about this too, but never pursued it before. Here is an educated guess based on dimensional analysis.
A pressure has the same units as an energy density:
$$
\mathrm{
\frac N{m^2} = \frac{N\,m}{m^3} = \frac J{m^3}
} $$
Electron capture takes kind of 1 MeV of energy, so maybe the transition becomes allowed once the energy density is something like 1 MeV/atom. Let's test this idea out: I vaguely remember that coal (or something) turns to diamond somewhere above one gigapascal:
\begin{align}
\mathrm{
1 \,GPa }&= \mathrm{10^9\frac J{m^3} \cdot \frac{1\,eV}{1.6\times10^{-19} \,J} \cdot \left(\frac{1\,m}{10^9\,nm}\right)^3 }\\
&\mathrm{= \frac 58\times10^{1}\,\frac{eV}{nm^3} }\\ &\mathrm{= 6 \,\frac{eV}{nm^3}
}\end{align}
Diamond is pure carbon, density 3.5 g/cm$^3$, so the volume of one atom is
\begin{align}
V &= \mathrm{ \frac{1\,cm^3}{3.5\,g} \cdot \frac{12\,g}{1\,mole}
\cdot \frac{1 \,mole}{6\times10^{23}\,atoms}\cdot \left(\frac{10^9\,nm}{100\,cm}\right)^3} \\
&\mathrm{ = \frac 2{3.5} \times10^{-2} \frac{nm^3}{atom} }\\
&\mathrm{ = 0.006 \frac{nm^3}{atom}
}\end{align}
Graphite (the initial state) has a density closer to 2 g/cm$^3$ so the volume of each atom is about 0.0035 nm$^3$.
This suggests that the transition from carbon to diamond at constant pressure takes an energy $E = P\Delta V \approx 0.015$ eV. This corresponds pretty closely to a tabulation of formation enthalpies which includes diamond:
creating diamond requires about 1.9 kJ/mole, or about 20 meV/atom. Shockingly close! I must have made a set of mistakes that cancel each other out :-)
Assuming this method is sound, the energy for electron capture (and other nuclear transitions) is generally about 1 MeV. So if your block of carbon kept the same density 0.006 nm$^3$/atom, and electron that disappears frees up 1/6 of the atom's volume, you'd need
$$
P = \frac E{\Delta V} = \mathrm{{10^6\,eV} \cdot \frac{6\,GPa\,nm^3}{1\,eV} \cdot \frac{6\, electrons/atom}{0.006\,nm^3/atom} = 6\times10^{9}\,GPa
}$$
Here's some company that thinks 15,000 bar = 1.5 GPa is a lot of pressure, so we have the sanity check that this estimate $P$ for the degeneracy pressure is "astronomical" (it's sixty trillion atmospheres).
I'm very interested to see if you get an answer from someone who isn't winging it.
Two years after I wrote this answer, Rob Jeffries adds in a comment
Neutronisation threshold in Carbon nuclei is 13.9 MeV. This is reached by degenerate electrons at pressure of about $10^{19}$ GPa.
So I guessed the neutronization energy to within a factor of ten, but my pressure is kind of embarrassingly off. But I explicitly made the assumption that the density of the diamond wouldn't change between ordinary conditions and the onset of neutron capture --- which of course isn't what happens.
In nature, the "overcoming" of electron degeneracy pressure occurs only in white dwarfs and degenerate stellar cores, where the density is as high as $10^7\rm\,g/cm^3$.
If you take my estimate $10^{10}\rm\,GPa$ above, increase the energy by a factor of ten, and decrease the volume per "atom" by a factor of $10^7$, you get within a factor of ten of RobJeffries's statement of the upper limit of electron degeneracy pressure in white dwarfs.