Let us name $M$ your matrix.
I am going to show that $\det(M)=0$.
You know that changing the order or rows and columns (said in matricial terms : pre- or post multiplying $M$ by permutation matrices) preserves the determinant up to a sign change. We can do such rows and columns permutations in such a way that the $5 \times 5$ block of zeros visible inside the matrix is moved in the lower-left corner.
We are now with the following matrix :
$$M'=\left( \begin{array}{c|c}A&B\\ \hline 0&D\end{array}\right)=$$
$$
=\left( \begin{array}{cccc|cccc}
*& * & * & * & * & * & * & *
\\
*& * & * & * & * & * & * & *
\\
*& * & * & * & * & * & * & *
\\0 & 0 & 0 & 0 & 0 & * & * & *\\
\hline
0 & 0 & 0 & 0 & 0 & * & * & *\\
0 & 0 & 0 & 0 & 0 & * & * & *\\
0 & 0 & 0 & 0 & 0 & * & * & *\\
0 & 0 & 0 & 0 & 0 & * & * & *
\end{array}\right)$$
with a $2 \times 2$ partition into blocks themselves $4 \times 4$. Due to the fact that the $C$ block is a block of zeros, we can conclude that :
$$\det(M')=\det(A)\det(D) \tag{1}$$
(see Remark below).
But $A$ has a row of zeros ; therefore $\det(A)=0$, and, as a consequence, $\det(M')=0$,... and $\det(M)=0$.
Remark : (1) is a known result, but one can deduce it from "your" rule,
$$\det(M')=\det(AD-BC)$$
being essential to twin this rule with the condition : iff $C$ and $D$ commute (see here) which is the case here because $C=0$. Then $$\det(AD-BC)=\det(AD)=\det(A)\det(D).$$
