Block matrices and their determinants

Flip the order of the Kronecker products to get $M'=A_n(I_n,I_n)+B_notimes T_n$, where $T_n=A_n(1,0)$. Note that $det M=det M'$. Since all blocks are polynomial in $A$, they commute, and therefore the determinant of $M'$ is $det(f(T_n))$, where $f(x)=det(A_n(1,1)+xB_n)$. That is, $f(x)=det(B_n)det(x+A_n(1,1)B_n)$ is $(-1)^{nchoose 2}$ times the characteristic polynomial of $H=-A_n(1,1)B_n$.

Let $t_n(x)$ be the characteristic polynomial of $T_n$. By repeated cofactor expansion on the first row, $t_n(x)=xt_{n-1}(x)+t_{n-2}(x)$. The initial conditions then imply that the $t_n$ is the $(n+1)^{th}$ Fibonacci polynomial. The roots of $t_n$ are $2icos(kpi/(n+1))$ for $k=1,dots,n$.

The eigenvalues of $H$ are worked out in "The eigenvalues of some anti-tridiagonal Hankel matrices". When $n$ is odd they are $1$ and $pmsqrt{3+2cos(frac{2kpi}{n+1})}$ for $k=1,dots,frac{n-1}{2}$. When $n$ is even they are $pmsqrt{1+4cos^2(frac{(2k+1)pi}{n+1})}$ for $k=0,dots,frac{n}{2}-1$.

By a quick diagonalization argument $det(f(T_n))$ is the resultant of $f$ and $t_n$. This plus some trig gives $$ det(M_{2n})=prod_{j,k=1}^nleft[1+4cos^2left(frac{jpi}{2n+1}right)+4cos^2left(frac{kpi}{2n+1}right)right]^2 $$ and $$ det(M_{2n-1})=prod_{j=1}^{n-1}left[1+4cos^2left(frac{jpi}{2n}right)right]^2prod_{k=1}^{n-1}left[1+4cos^2left(frac{jpi}{2n}right)+4cos^2left(frac{kpi}{2n}right)right]^2. $$