An attempt to prove conjecture 1. It does not give the correct answer, I am posting it in order to, either, soon correct it, or to stand corrected.
It is helpful to represent the determinant of an $n\times n$ matrix $M$ as an integral over anticommuting (Grassmann) variables $\theta=(\theta_1,\theta_2,\ldots\theta_n)$, and their conjugates $\bar{\theta}=(\bar{\theta}_1,\bar{\theta}_2,\ldots\bar{\theta}_n)$,
$$\det M=\int d\theta\int d\bar{\theta}\,e^{\bar{\theta}\cdot M\cdot\theta}=\int d\theta\int d\bar{\theta}\,\prod_{i=1}^n\left(1+\bar{\theta}_i\sum_{j=1}^n M_{ij}\theta_j\right),\tag{1}$$ as explained, for example, in these lecture notes.
Apply this to $M=X^2+Y^2$, $$\det(X^2+Y^2)=\int d\theta\int d\bar{\theta}\,e^{\bar{\theta}\cdot X^2\cdot\theta}e^{\bar{\theta}\cdot Y^2\cdot\theta}$$ $$\qquad=\int d\theta\int d\bar{\theta}\,\prod_{i,i'=1}^n\left(1+\bar{\theta}_i\sum_{j,k=1}^n X_{ik}X_{kj}\theta_j\right)\left(1+\bar{\theta}_{i'}\sum_{j',k'=1}^n Y_{i'k'}Y_{k'j'}\theta_{j'}\right).\tag{2}$$ We can now directly take the expectation value over the independent normally distributed matrix elements of $X$ and $Y$. Only the terms $i=j=k$ and $i'=j'=k'$ survive the average, $$\mathbb{E}[\det(X^2+Y^2)]=\int d\theta\int d\bar{\theta}\,\prod_{i=1}^n\left(1+\bar{\theta}_i\theta_i\right)^2=\int d\theta\int d\bar{\theta}\,\prod_{i}^n\left(1+2\bar{\theta}_i\theta_i\right)$$ $$\qquad\qquad=2^n.\tag{3}$$ Not the desired answer $(n+1)!$ --- need to correct it.