Introduce the $2^{n-1}$-dimensional matrices $B_n$ 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 $2^{n-1}$-dimensional 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 $\det(B_n)$?