The question I have is quite specific. So in the hope that this post might help others in the future, my problem boils down to solving the following integral:

$$I_\tau=\int \prod_{i, j=1}^{N} d J_{i j} \exp  \left\{-\frac{N}{2(1-\tau^2)}\left( \sum_{i, j, k} J_{k i} A_{i j} J_{k j}-\tau\sum_{ij}J_{ij}J_{ji}\right)\right\}$$
Assuming we can compute the determinant of $A$, is there a general identity to compute $I_\tau$?

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Simplifying their argument, in the following paper [1], the authors compute the following Gaussian integral over all $N\times N$ real asymmetric matrices $J_{ij}$: 

$$I=\int \prod_{i, j=1}^{N}d J_{i j} \exp \left\{-\frac{N}{2} \sum_{i, j, k} J_{k i} A_{i j} J_{k j}\right\}$$

Where $A_{i j}=\frac{z_{i}^{*} z_{j}}{N}+\frac{z_{j}^{*} z_{i}}{N}+\delta_{i j}$ and $z_i\in \mathbb{C}$.

If I define the vector $\mathbf{x}$ such that:
$$\mathbf{x}:=\left(\begin{array}{c}
J_{11} \\
J_{12} \\
\vdots \\
J_{1 n} \\
J_{21} \\
J_{22} \\
\vdots \\
J_{n n}
\end{array}\right) \in \mathbb{R}^{n^{2}}$$
$$\implies x_{N(i-1))+j}=J_{ij}$$
Then we can represent $I$ as the following integral:
$$I=\int \left(\prod_{i, j=1}^{N} \mathrm{d}x_{N(i-1))+j} \right)\exp \left(-\frac{N}{2} \mathbf{x}^T\Sigma \mathbf{x}\right)$$
Where $\Sigma$ simply is:
$$\Sigma=\left(\begin{array}{cccc}
\mathbf{A} & 0 & \cdots & 0 \\
0 & \mathbf{A} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \mathbf{A}
\end{array}\right)$$

Thus $I\propto \frac{1}{\sqrt{|\Sigma|}}=|A|^\frac{-N}{2}$.

Defining $r\equiv\frac{1}{N}\sum_i |z_i|^2$, the authors observe that to order $1/\sqrt{N}$, the matrix $A$ has all but two of its eigenvalues equal to one. The two exceptions are the eigenvectors $z_i$ and $z^*_i$ , both with eigenvalue $(1+r)$. This observation allows them to compute the determinant of $\Sigma$ and thus the Gaussian integral $I$.

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Now, considering the integral $I_\tau$ using the same method:
$$I_\tau=\int \left(\prod_{i j} \mathrm{d} J_{i j}\right)\exp  \left\{-\frac{N}{2(1-\tau^2)}\left( \sum_{i, j, k} J_{k i} A_{i j} J_{k j}-\tau\sum_{ij}J_{ij}J_{ji}\right)\right\}$$
$$I_\tau=\int\left(\prod_{i j} \mathrm{d} J_{i j}\right) \exp \left\{-\frac{N}{2(1-\tau^2)} \sum_{i j, k, l} J_{i j}\left(A_{j l} \delta_{i k}-\tau\delta_{i l} \delta_{j k}\right) J_{k l}\right\}$$
$$\implies I_\tau=\int\left(\prod_{i, j=1}^{N} \mathrm{d} x_{N(i-1))+j}\right) \exp \left(-\frac{N}{2} \mathbf{x}^{T} \Sigma \mathbf{x}\right)$$
Where this time:
$$\Sigma_{n(i-1)+j,n(k-1)+l}=A_{j l} \delta_{i k}-\tau\delta_{i l} \delta_{j k}  \quad \forall i, j, k, l \in \mathbb{N} \cap[1, N]$$

For large $N$, how can we compute the determinant of $\Sigma$ this time? It seems to be a non-trivial question.

Any remark or comment is always appreciated, thank you.

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[1]: Rajan, K., & Abbott, L. F. (2006). Eigenvalue spectra of random matrices for neural networks. Physical review letters, 97(18), 188104.

Link: https://www.researchgate.net/profile/Kanaka_Rajan/publication/6643146_Eigenvalue_Spectra_of_Random_Matrices_for_Neural_Networks/links/0912f50f7347f4863b000000.pdf