What's the Q factor of a digital filter's pole?
it depends on how you map the analog filter to the digital filter and how the s-plane poles get mapped to the z-plane poles (ya know, the "bilinear transform" vs. "impulse invariant" vs. whatever else is out there).  

probably, if it were up to me, for the purpose of defining $Q$, i would map the poles as you would map $s$ to $z$ with

$$ z = e^{sT} $$

where $T \triangleq \frac{1}{f_\text{s}}$ is the sampling period and $f_\text{s}$ is the sample rate.  that would be assuming impulse invariant.  the damping rate of the digital filter would be the same as the analog filter because both have the same exponential impulse responses (it's just that the digital filter's impulse response is the sampled analog impulse response).

for bilinear transform the $s$ to $z$ mapping is

$$ z = \frac{1 + sT/2}{1 - sT/2} $$

if you're using $Q$ to define the degree of resonance (like the dB boost you get around the resonant frequency, whether the filter is LPF, BPF or HPF), then this mapping will preserve the height of that bump for a given $Q$.

i dunno if another mapping would be useful for this definition of digital Q.  in the Audio EQ Cookbook, i worried a little about mapping BPF bandwidth from s-plane to z-plane and tossed in a fudge factor because bilinear transform sorta scrunches the bandwidth as the resonant frequency gets closer to Nyquist, $f_\text{s}/2$.  if preservation of bandwidth, as you map from s-plane to z-plane, is important to you, then i would take a look at that.

so, given the s-plane pole location (which is a function of $Q$), map it to the z-plane and you'll get an expression for $Q$ based on the real and imaginary parts (or magnitude and angle) of the z-plane poles.