This is easy to do using a graphical calculus for contractions of old-fashioned tensors. See this recent article for an example of application of such techniques and hopefully useful references.
Here is a proof of Conjecture 1. When I have time I will try to make nicer pictures.
Then the determinant is given by $$ {\rm det}(M)= \ \ \begin{array}{ccccccc} & & --& -- &-- & & \\ & / & --&-- & --&\backslash & \\ / & / & & & & \backslash & \backslash \\ | & | & & & & | & | \\ | & | & & & & M & M \\ | & | & & & & | & | \\ | & | & & & & = & = \\ | & | & & & & | & | \\ \backslash & \backslash & & & &/ & / \\ & \backslash & --& -- &-- &/ & \\ & &-- & -- &-- & & \end{array} $$$$ {\rm det}(M)= \ \ \begin{array}{ccccccc} & & --& -- &-- & & \\ & / & --&-- & --&\backslash & \\ / & / & & & & \backslash & \backslash \\ | & | & & & & | & | \\ | & | & & & & M & M \\ | & | & & & & | & | \\ | & | & & & & = & = \\ | & | & & & & | & | \\ \backslash & \backslash & & & & / & / \\ & \backslash & --& -- &-- & / & \\ & & -- & -- &-- & & \end{array} $$
Now take $M=X^2+Y^2$ and expand by multilinearity. That means that, in the previous picture, each $$ \begin{array}{c} | \\ M \\ | \end{array} $$ becomes $$ \begin{array}{c} | \\ X \\ | \\ X \\ | \end{array} $$ or $$ \begin{array}{c} | \\ Y \\ | \\ Y \\ | \end{array} $$ Now the placement of the pairs of $X$'s and $Y$'s dodoes not matter, because exchanging positions creates two twists, one on top and one below the antisymmetrizer. Undoing them gives a factor $(-1)^2$, i.e., does nothing. So we get a sum $$ \sum_{n_1+n_2=n}\frac{n!}{n_1! n_2!}\times $$ the expectation of the picture above with the $M$'s (now $n$ of them) are replaced by something like $$ \begin{array}{} | & | & \cdots &|&|&|& \cdots&|& \\ X& X &\cdots& X &Y& Y&\cdots& Y \\ | & | & \cdots &|&|& |&\cdots&|& \\ X& X &\cdots& X &Y& Y&\cdots& Y \\ | & | & \cdots &|&|&|&\cdots&|& \end{array} $$$$ \begin{array}{cccccccc} | & | & \cdots & | & | & | & \cdots & | \\ X & X & \cdots & X & Y & Y & \cdots & Y \\ | & | & \cdots & | & | & | & \cdots & | \\ X & X & \cdots & X & Y & Y & \cdots & Y \\ | & | & \cdots & | & | & | & \cdots & | \end{array} $$ where there are $n_1$ vertical strands of $X$'s followed by $n_2$ strands of $Y$'s.
Now we apply the Isserlis-Wick Theorem in order to perform the Gaussian integrals, as a sum over perfect matchings of $X$'s among themselves and of $Y$'s among themselves. The main ingredient is the graphical identity $$ \mathbb{E}\left[ \begin{array}{c} | \\ X \\ | \end{array} \begin{array}{c} | \\ X \\ | \end{array} \right]= \begin{array}{} | & & | \\ - & -- & - \\ & & \\ - & -- & - \\ | & & | \end{array} $$ namely, with indices, $$ \mathbb{E}[X_{ij}X_{k\ell}]=\delta_{ik}\delta_{j\ell} $$ A better picture than using "$--$"'s and "$|$" would show a cup $\bigcup$ on top and a cap $\bigcap$ on the bottom.