Expanding the determinant along the third row,

$\det =-5(ad-bc).$

So the matrix is singular iff $ad=bc$, with each of $a,b,c,d$ chosen from an $8$-element set. We count the singular matrices.

Case 1: $ad=bc=0$. The number of pairs $(a,d)$ with $ad=0$ is $8^2-7^2=15$. Similarly $15$ for $(b,c)$. Total $15\times 15=225$.

Case 2: $ad=bc\ne 0$. Then $a,b,c,d$ all lie in the $7$-element nonzero set. Write $ad=bc=N$. For each unordered factorization $N=xy$ with $x,y$ in the nonzero set, the pair $(a,d)$ has $1$ ordering if $x=y$ and $2$ orderings if $x\ne y$; same for $(b,c)$.

Sub-case 2a, $a=d$: this contributes $7$ choices of common value, with $(b,c)$ being the same square, giving $7\times 1=7$.

Sub-case 2b, $a\ne d$ and the pair is $\{x,y\}$ with $x\ne y$: there are $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ unordered pairs, and the ordered count is $21\times 2\times 2=84$.

Total Case 2: $7+84=91$. Singular matrices: $225+91=316$.

Invertible matrices: $8^4-316=4096-316=3780.$