Yes and No. On the one hand, you can prove by a simple induction that $B_n$ commuttes with $J_n$. Now, $B_n$ appears to be given blockwise, where all the blocks commutte to each other. This allows you to write that the determinant of $B_n$ is the ``determinant of determinants'' (this is false for non-commutting blocks): $$\det B_n=(\det B_{n-1})^2-(\det b_nJ_{n-1})^2=(\det B_{n-1})^2-b_n^{2^{n-1}}.$$ On the other hand, it seems difficult to exploit this relation to give a closed form for $\det B_n$. Recall that even the iteration $z\mapsto z^2-a$ has an outstanding complexity, which yields Julia and Fatou sets.

Yes and No. On the one hand, you can prove by a simple induction that $B_n$ commutes with $J_n$. Now, $B_n$ appears to be given blockwise, where all the blocks commute with each other. This allows you to write that the determinant of $B_n$ is the ``determinant of determinants'' (this is false for non-commuting blocks): $$\det B_n=(\det B_{n-1})^2-(\det b_nJ_{n-1})^2=(\det B_{n-1})^2-b_n^{2^{n-1}}.$$ On the other hand, it seems difficult to exploit this relation to give a closed form for $\det B_n$. Recall that even the iteration $z\mapsto z^2-a$ has outstanding complexity, which yields Julia and Fatou sets.