How to calculate the delay generated by a 2nd order all-pass filter
The phase of a second-order all-pass system with poles at $z=re^{j\theta}$ and $z=re^{-j\theta}$ is $[1]$

$$\phi(\omega)=-2\omega-2\arctan\left[\frac{r\sin(\omega-\theta)}{1-r\cos(\omega-\theta)}\right]-2\arctan\left[\frac{r\sin(\omega+\theta)}{1-r\cos(\omega+\theta)}\right]\tag{1}$$

The poles are the roots of the denominator polynomial $z^2+a_1z+a_2$. For the given values of $a_1$ and $a_2$ you get $r=0.93264$ and $\theta=0.040271$.

The delay of a sinusoid with frequency $\omega_0$ when passing through the all-pass filter is given by the phase delay

$$\tau_p=-\frac{\phi(\omega_0)}{\omega_0}\tag{2}$$

Evaluating $(2)$ using $(1)$ with $\omega_0=2\pi\frac{262.5}{44100}$ gives a phase delay of $42.7$ samples, which is pretty close to the value you've determined experimentally.



$[1]$ Oppenheim, Schafer, Buck, Discrete-Time Signal Processing, 2nd ed., p. 275.