Consider two sequences of random numbers
\begin{align} A &= \{a_0, a_1, \ldots a_N\} \\ B &= \{b_0, b_1, \ldots b_N\} \, . \end{align}
where each $a$ and $b$ value is independently drawn from a Gaussian distribution
$$G_\sigma(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left[ -\frac{x^2}{2 \sigma^2}\right] \, .$$
Suppose we compute the discrete Fourier transform of each sequence
\begin{align} \tilde{A}_k &\equiv \sum_{n=0}^{N-1} a_n \exp\left[-i 2 \pi n k / N \right] \\ \tilde{B}_k &\equiv \sum_{n=0}^{N-1} b_n \exp\left[-i 2 \pi n k / N \right] \, . \end{align}
What is the statistical distribution of
$$\tilde{C}_k \equiv \tilde{A}_k \tilde{B}_k^* \, ?$$
Attempt
Using the fact that the distribution of a sum of random variables is the convolution of the distributions of the things being summed, we can find that $\tilde{A}_k$ is a complex number whose real and imaginary parts are independent and both Gaussian distributed with standard deviation $\sigma \sqrt{2N}$. Then, using the rules for transformation of random variables (and the convolution rule again) we can find that the distribution of the square of the radius
$$r^2 \equiv (\text{Re} \tilde{A}_k)^2 + (\text{Im} \tilde{A}_k)^2$$
is
$$P_{r^2}(\alpha) = \frac{1}{\sigma^2 N} \exp \left[ -\alpha/\sigma^2 N \right] \qquad (\alpha > 0) \, .$$
Again using transformation rules we find the distribution of the radius
$$P_r(r) = \left( \frac{2}{\sigma^2 N} \right) r \exp\left[-r^2 / \sigma^2 N \right] \qquad (r > 0) \, .$$
Note that from this we can compute $\langle r \rangle = \sigma \sqrt{\pi N} / 2$.
The statistics of $\tilde{B}_k$ are obviously the same as the statistics of $\tilde{A}_k$.
Now we consider $\tilde{C}_k \equiv \tilde{A}_k \tilde{B}_k$. The phase of $\tilde{C}_k$ is the sum of the phases of $\tilde{A}_k$ and $\tilde{B}_k$. As the phases of $\tilde{A}_k$ and $\tilde{B}_k$ are both uniformly distributed over $[0,\pi)$, it's obvious that the phase of $\tilde{C}_k$ is also uniformly distributed over that range. The radius is the interesting part. The radius of the product is the product of the radii. Fortunately, we know that the product of two random variables $Z=XY$ follows the distribution
$$P_Z(z) = \int_{-\infty}^\infty P_X(x) P_Y(z/x) \frac{1}{|x|}dx \, .$$
Therefore, the distribution of the radius of $\tilde{C}_k$ is
\begin{align} P_{|\tilde{C}_k|}(z) &= \int_{-\infty}^\infty P_r(r) P_r(z/r) \frac{dr}{|r|} \\ &= \int_0^\infty P_r(r) P_r(z/r) \frac{dr}{r} \\ &= \left( \frac{2}{\sigma^2 N} \right)^2 z \int_0^\infty \exp \left[ - \frac{ r^2 + (z/r)^2 }{\sigma^2 N} \right] \frac{dr}{r} \, . \end{align}
I have not been able to solve this integral. However, we can find the average value of $|\tilde{C}_k|$: \begin{align} \langle |\tilde{C}_k| \rangle &= \int_0^\infty z P_{|\tilde{C}_k|}(z) \, dz \\ &= \left( \frac{2}{\sigma^2 N}\right)^2 \int_0^\infty \int_0^\infty dz \, dr \, \frac{z^2}{r} \exp \left(-\left(r^2 + (z/r)^2 \right)/(\sigma^2 N) \right) \\ &= \left( \frac{2}{\sigma^2 N}\right)^2 \frac{\sqrt{\pi}}{4} (\sigma^2 N)^{3/2} \int_0^\infty dr \, r^2 \exp \left( -r^2 / \sigma^2 N \right) \\ &= \frac{\pi}{4} \sigma^2 N \, . \end{align} which, interestingly, is exactly the square of $\langle r \rangle$ for either $\tilde{A}_k$ or $\tilde{B}_k$. This is better than nothing, but the full distribution remains to be found.
Not a very fun answer, but your integral can be solved explicitly by using e.g. Mathematica. Given the following input:
The computer spits out the explicit solution: Simply a Bessel function. The final result is
$$ P(z)=\bigg(\frac{2}{\sigma^2N}\bigg)^2 z \cdot \underbrace{K_0\bigg(\frac{2z}{N\sigma^2}\bigg)}_{\text{integral}}$$
Where $K_0$ is defined e.g. on this page. One particularly simple representation for it is
$$K_0(z)=\int_0^\infty \cos(z\sinh t)\mathrm d t $$
Perhaps, one can actually construct this solution by hand once the answer is known, but I'll leave that as a proverbial exercise to the reader.