It all depends on the question. For most situations I would not consider this from the power standpoint as was done in http://personal.health.usf.edu/ywu/logistic.pdf. Rather model reliability (e.g., calibration) is often a more appropriate consideration. For that purpose, start by considering simple cases with no competing covariates.
Suppose there were no covariates, so that the only parameter in the
model were the intercept. What is the sample size required to allow
the estimate of the intercept to be precise enough so that the
predicted probability is within 0.1 of the true probability with 0.95
confidence, when the true intercept is in the neighborhood of zero?
The answer is n=96. What if there were one covariate, and it was
binary with a prevalence of $\frac{1}{2}$? One would need 96 subjects
with $X=0$ and 96 with $X=1$ to have an upper bound on the margin of
error for estimating $\textrm{Prob}\{Y=1|X=x\}$ not exceed 0.1 for
either value of $x$. The general formula for the sample size
required to achieve a margin of error of $\delta$ in estimating a
true probability of $\theta$ at the 0.95 confidence level is $n =
(\frac{1.96}{\delta})^{2} \times \theta(1 - \theta)$. Set $\theta =
\frac{1}{2}$ (intercept=0) for the worst case.
This idea can be extended by simulation of a number of continuous covariates or by imagining the effective number of unique orthogonal "groups" there are in the covariate space for which one desires good prediction accuracy. $96\times k$ where $k$ is the number of "groups" would be an estimate of the required sample size.