Is the phase of a digital filter always an odd function from $[-\pi, \pi]$
To expand on RBJ's comment from above, your statement is true iff the impulse response of the filter is real. Recall that the frequency response of a filter is the discrete time Fourier transform of its impulse response. That is:

$$
X(\omega) = \sum_{n=-\infty}^{\infty} x[n] e^{j\omega n}
$$

Look at the relationship between $X(\omega)$ and $X(-\omega)$:

$$
X(-\omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n}
$$
$$
X(-\omega) = \sum_{n=-\infty}^{\infty} x[n] \left(e^{j\omega n}\right)^*
$$

If $x[n]$ is real, then we can absorb it into the conjugate:

$$
X(-\omega) = \sum_{n=-\infty}^{\infty} \left(x[n] e^{j\omega n}\right)^*
$$

And since conjugation is a linear operation, we can pull it around the sum as well:

$$
X(-\omega) = \left(\sum_{n=-\infty}^{\infty} x[n] e^{j\omega n}\right)^*
$$
$$
X(-\omega) = \left(X(\omega)\right)^*
$$

So $X(\omega)$ exhibits Hermitian symmetry; its phase is oddly symmetric about $\omega = 0$, while its magnitude is evenly symmetric.