Introduce the $2^{n-1}\times 2^{n-1}$ matrix $B_n$ recursively as follows: $B_1(b_1)=\begin{pmatrix} b_1\end{pmatrix}$ and $$B_n(b_1,\dots,b_n)=\begin{pmatrix} B_{n-1}(b_1,\dots,b_{n-1})& b_nJ_{n-1}\\ b_nJ_{n-1}&B_{n-1}(b_1,\dots,b_{n-1}) \end{pmatrix}.$$ Here $J_n$ is a $2^{n-1}\times 2^{n-1}$ matrix with $1$'s on the antidiagonal and zeros elsewhere.

Example. For $n=2$ and $n=3$, we have $$B_2(b_1,b_2)=\begin{pmatrix} b_1&b_2\\b_2&b_1\end{pmatrix} \qquad\text{and} \qquad B_3(b_1,b_2,b_3)=\begin{pmatrix} b_1&b_2&0&b_3\\b_2&b_1&b_3&0\\0&b_3&b_1&b_2\\b_3&0&b_2&b_1\end{pmatrix}.$$

Question. Is there a closed (or interesting) formula for the determinant $\det(B_n)$?