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This does not seem to be an easy road to an answer, I will leave it for the record (or delete it if so requested)

For conjecture 1, 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 now take the expectation value over the independent normally distributed matrix elements of $X$ and $Y$, $$\mathbb{E}[\det(X^2+Y^2)]=\int d\theta\int d\bar{\theta}\,\left(\mathbb{E}\biggl[\prod_{i=1}^n\biggl(1+\bar{\theta}_i\sum_{j,k=1}^n X_{ik}X_{kj}\theta_j\biggr)\biggr]\right)^2.\tag{3}$$

So I need to evaluate a Gaussian average of the form $$Z_n=\mathbb{E}\biggl[\prod_{i=1}^n\biggl(1+\sum_{j,k=1}^n X_{ik}X_{kj}c_{ij}\biggr)\biggr]$$ and then perform the remaining integral of $Z_n^2$ over the coefficients $c_{ij}=\bar{\theta}_i\theta_j$, with the help of the identities $$\int d\theta_id\bar{\theta}_i=0,\;\;\int d\theta_id\bar{\theta}_i\,\bar\theta_i=0,\;\int d\theta_id\bar{\theta}_i\,\bar\theta_i\theta_i=1.$$