Presumably you mean to say that $|c_{ij}|\le B$, since if you just say $c_{ij}<B$, then the $c_{ij}$ could be very negative.

But even then your bound $B^n$ is false. Here's a counterexample: $B=\begin{pmatrix} 2&1\\-1&2\\\end{pmatrix}$, then entries are bounded by $2$, but $\Delta=5>2^2$. The most trivial estimate is that $\Delta$ is a sum of $n!$ terms, each of which is at most $B^n$, so the trivial bound is $n!B^n$. One can, of course, do much better. Hadamard's inequality gives $|\Delta|\le n^{n/2}B^n$.