Allpass group delay formula and effect on Karplus-Strong
For the Karplus-Strong algorithm, your primary concern is for phase delay,
$$ \tau_\phi(\omega) \ = \ - \frac{\phi(\omega)}{\omega} $$

rather than group delay.
$$ \tau_g(\omega) \ = \ - \frac{d \phi(\omega)}{d \omega} $$

where
$$ \phi(\omega) \ \triangleq \ \arg \left\{ H(e^{j \omega}) \right\} $$
and $H(z)$ is the transfer function of your filter, and
$$  H(e^{i \omega}) \ = \ \left| H(e^{j \omega}) \right| e^{j \phi(\omega)} $$

For phase delay, it is important to unwrap the phase, which is not necessary for 1st-order APFs.

This is because, for a sinusoid, it's the phase delay that is the actual delay, in units of time, that the filter delays a simple sinusoid.  The group delay is the amount the envelope of the sinusoid is delayed.

For linear-phase filters, $\tau_\phi(\omega)=\tau_g(\omega)$ for all $\omega$ and for any filter $\tau_\phi(0)=\tau_g(0)$.  Phase delay and group delay are not generally the same value in other circumstances.

Now for the 1st-order APF with its pole at $p$:
$$ H(z) \ = \ \frac{1-pz}{z-p}$$
the phase delay in sample units is
$$ \tau_\phi(\omega) \ = \ \frac{1}{\omega} \arctan\left(\frac{(1-p^2)\sin(\omega)}{(1+p^2)\cos(\omega) - 2p} \right) $$
and the group delay in sample units is (same expression MattL gets)
$$ \tau_g(\omega) \ = \ \frac{1-p^2}{1+p^2 - 2p\cos(\omega)} $$

the delay (phase or group) at DC is $\tau_\phi(0) = \tau_g(0) = \frac{1+p}{1-p}$ and we normally have $-1 \le p \le 0$ to get a delay of between 0 and 1 sample.

Now, in the Karplus-Strong algorithm, the frequency of every harmonic (perhaps better called a "partial"), including the 1st harmonic (which we'll call the "fundamental") is the reciprocal of the total loop delay (phase delay) at the frequency of that partial.  That is a circular mathematical relationship.  For the $k$th partial (and assuming normalized sampling frequency $f_s=1$):
$$ f_k \ = \ \frac{k}{n + \tau_\phi(2 \pi f_k)} $$
where $n$ are the integer number of samples in the delay line that is in series with the APF.  Given your fundamental frequency $f_1$, the value of $n$ is $$ n \ = \ \left\lfloor \frac{1}{f_1} \right\rfloor \ = \ \mbox{floor}\left(\frac{1}{f_1}\right) $$
 (Be sure to add 1 sample that is necessary in a digital filter with feedback.)

Now, often we approximate using the DC delay, but to design this exactly, you need to somehow solve that frequency-dependent phase delay at the frequency of each partial.  That's a bitch.  So you would need to solve
$$ \frac{k}{f_k} \ = \ n + \tau_\phi(2 \pi f_k) \ = \ n + \frac{1}{2 \pi f_k} \arctan\left(\frac{(1-p^2)\sin(2 \pi f_k)}{(1+p^2)\cos(2 \pi f_k) - 2p} \right) $$
for $f_k$ given every integer $k$.