Expanding on my comment, $\{\phi(\tau)\}$ is a non-stationary nonGaussian random process, and I doubt that there is any simple answer (or even a rather complicated one) for the probability density function of the random variable
$\phi(\tau)$ for an arbitrary value of $\tau$. But, the (time-varying) mean
function of the process is easy to calculate. We have
$$\begin{align*}
E[\phi(\tau)] &= E\left[\int_{-T/2}^{T/2} g(t)g(t-D+\tau)\, \mathrm dt\right]\\
&= \int_{-T/2}^{T/2} E[g(t)g(t-D+\tau)]\, \mathrm dt\\
&= \int_{-T/2}^{T/2} R_g(\tau-D)\, \mathrm dt\\
&= T\cdot R_g(\tau-D)
\end{align*}$$
where $R_g(\cdot)$ is the autocorrelation function of the input process
$\{g(t)\}$. Note that it is not necessary that the input process
be Gaussian for this to hold; wide-sense-stationarity is enough.
Since autocorrelation functions have a peak at the origin,
we see that $\phi(D)$ has
the largest mean value. Also, the mean value decays away symmetrically
about $D$: that is, $E[\phi(D+\epsilon)] = E[\phi(D-\epsilon)]$
and
$$|E[\phi(\tau)]| \leq E[\phi(D)] = T\cdot R_g(0).$$
Finding the variance of $\phi(\tau)$ is a much messier calculation
that may or may not be included in the paper cited by
Charna (which is behind a paywall).
