This is the translation of the @AbdelmalekAbdesselam answer to the standard algebraic notation, for those who, like me, feel not so comfortable with birdtracks notation. The main idea is to expand $\det(X^2 + Y^2)$ in the sum of monomials on variables $X_{ij},Y_{ij}$ and then apply Isserlis' formula. Namely, one has $$\mathbb E[\det(X^2+Y^2)] = \sum_{\sigma \in \Sigma_n} (-1)^{\sigma} \mathbb E[(a_{1,\sigma 1} + b_{1,\sigma 1}) ... (a_{n,\sigma n} + b_{n,\sigma n})] =$$$$\mathbb E[\det(X^2+Y^2)] = \sum_{\sigma \in \Sigma_n} (-1)^{\sigma} \mathbb E[(a_{1,\sigma 1} + b_{1,\sigma 1}) \cdots (a_{n,\sigma n} + b_{n,\sigma n})] =$$ where $A = X^2, B=Y^2$. Now, after expanding and regrouping $a$-terms with b-terms and using various symmetries of the problem $$=\sum_{\sigma \in \Sigma_n}\sum_{\substack{n_1 + n_2 = n \\\{i_1,...,i_{n_1}\} \sqcup \{j_1,...,j_{n_2}\} = \{1,...,n\} \\ i_1 < ... < i_{n_1}\\j_1 < ... < j_{n_2}}} (-1)^{\sigma} \mathbb E[a_{i_1,\sigma i_1} ... a_{i_{n_1}, \sigma i_{n_1}}]\mathbb E[b_{j_1,\sigma j_1} ... b_{j_{n_2}, \sigma j_{n_2}}] =$$$$=\sum_{\sigma \in \Sigma_n}\sum_{\substack{n_1 + n_2 = n \\\{i_1,\ldots,i_{n_1}\} \sqcup \{j_1,\ldots,j_{n_2}\} = \{1,\ldots,n\} \\ i_1 < \cdots < i_{n_1}\\j_1 < \cdots < j_{n_2}}} (-1)^{\sigma} \mathbb E[a_{i_1,\sigma i_1} \cdots a_{i_{n_1}, \sigma i_{n_1}}]\mathbb E[b_{j_1,\sigma j_1} \cdots b_{j_{n_2}, \sigma j_{n_2}}] =$$ $$=\sum_{n_1+n_2=n}\frac{n!}{n_1! n_2!} \sum_{\sigma_1 \in \Sigma_{n_1}, \sigma_2 \in \Sigma_{n_2}} (-1)^{\sigma_1} (-1)^{\sigma_2} \mathbb E[a_{1,\sigma_1 1} ... a_{n_1, \sigma n_1}]\mathbb E[b_{1,\sigma_2 1} ... b_{n_2, \sigma_2 n_2}] =$$$$=\sum_{n_1+n_2=n}\frac{n!}{n_1! n_2!} \sum_{\sigma_1 \in \Sigma_{n_1}, \sigma_2 \in \Sigma_{n_2}} (-1)^{\sigma_1} (-1)^{\sigma_2} \mathbb E[a_{1,\sigma_1 1} \cdots a_{n_1, \sigma n_1}]\mathbb E[b_{1,\sigma_2 1} \cdots b_{n_2, \sigma_2 n_2}] =$$ $$=\sum_{n_1 + n_2 = n} \frac{n!}{n_1! n_2!} \mathbb E[\det(Z_{n_1}^2)] \mathbb E[\det(Z_{n_2}^2)]$$
Wherewhere $Z_{r}$$Z_r$ is an $r \times r$ matrix filled with iid normal distributions. So it is enough to prove
$$\mathbb E[\det(Z_{r})^2]=r!$$ Via Isserlis' formula only quadratic monomials of $\det(Z_{r})^2$ will contribute to the answer, and there are exactly $r!$ of them.
