This is a fair example of the following theorem : let $A_{ij}\in M_r(k)$ be pairwise commuting matrices for $1\le i,j\le d$, and let $A\in M_{dr}(k)$ be the matrix whose $r\times r$ blocks are the $A_{ij}$'s. Then $\det A$ equals the determinant of the matrix $B\in M_r(k)$ obtained by computing the formal determinant of the blocks. Example :
$$\det\begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}=\det(A_{11}A_{22}-A_{12}A_{21}).$$
Mind that the formula is false if the blocks don't commute.

In your case, that means that
$$\det A=\det P_N(J_n),$$
where $P_N(X)$ is the determinant of the tridiagonal matrix whose diagonal entries are $X$ and the sub/super-diagonal entries are ones. This is the monic polynomial whose roots are the numbers $2\cos\frac{k\pi}{N+1}$, $1\le k\le N$.

In particular, the eigenvalues of $J_n$ are the numbers $1+2\cos\frac{j\pi}{n+1}\,$. Hence the formula
$$\det A=\prod_{j=1}^nP_N\left(1+2\cos\frac{j\pi}{n+1}\right).$$