Finally, here is an argument that I like to present.

We proceed with induction on $n$. The base case $n=2$ is obvious. So assume $n\geq3$. Denote $\widehat{A_n}=A_nJ_n, \widehat{B_n}=B_nJ_n$ and let $I_n$ be the $2^{n-1}$-dimensional identity matrix. It is immediate that $$\widehat{A_n}=\begin{pmatrix} a_nI_{n-1}&\widehat{A}_{n-1}\\ \widehat{A}_{n-1}&-a_nI_{n-1} \end{pmatrix}.$$ An important difference surfaces: now the two bottom pair of block-matrices \it commute. \rm As a result, $$\begin{align} \det A_n&=\det J_n\det\widehat{A}_n=\det(-a_n^2I_{n-1}-\widehat{A}_n^2)=(-1)^{2^{n-2}}\det(a_nI_{n-1}\pm i\widehat{A}_n).\end{align}$$ The proof follows. $\square$

Finally, here is an argument that I came up with.

We proceed with induction on $n$. The base case $n=2$ is obvious. So assume $n\geq3$. Denote $\widehat{A_n}=A_nJ_n, \widehat{B_n}=B_nJ_n$ and let $I_n$ be the $2^{n-1}$-dimensional identity matrix. It is immediate that $$\widehat{A_n}=\begin{pmatrix} a_nI_{n-1}&\widehat{A}_{n-1}\\ \widehat{A}_{n-1}&-a_nI_{n-1} \end{pmatrix}.$$ An important difference surfaces: now the two bottom pair of block-matrices \it commute. \rm As a result, $$\begin{align} \det A_n&=\det J_n\det\widehat{A}_n=\det(-a_n^2I_{n-1}-\widehat{A}_n^2)=(-1)^{2^{n-2}}\det(a_nI_{n-1}\pm i\widehat{A}_n).\end{align}$$ The proof follows. $\square$