For $n\in\Bbb{N}$, define three matrices $A_n(x,y), B_n$ and $M_n$ as follows:

(a) the $n\times n$ tridiagonal matrix $A_n(x,y)$ with main diagonal all $y$'s, superdiagonal all $x$'s and subdiagonal all $-x$'s. For example, $$ A_4(x,y)=\begin{pmatrix} y&x&0&0\\-x&y&x&0\\0&-x&y&x \\0&0&-x&y\end{pmatrix}. $$

(b) the $n\times n$ antidigonal matrix $B_n$ consisting of all $1$'s. For example, $$B_4=\begin{pmatrix} 0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}.$$

(c) the $n^2\times n^2$ block-matrix $M_n=A_n(B_n,A_n(1,1))$ or using the Kronecker product $M_n=A_n(1,0)\otimes B_n+I_n\otimes A_n(1,1)$.

Question.What is the determinant of $M_n$?

**UPDATE.** For even indices, I conjecture that

$$\det(M_{2n})=\prod_{j,k=1}^n\left[1+4\cos^2\left(\frac{j\pi}{2n+1}\right)+4\cos^2\left(\frac{k\pi}{2n+1}\right)\right]^2.$$

This would confirm what Philipp Lampe's "perfect square" claim.