A Hadamard matrix $H$ is a matrix with entries $\pm1$ and orthogonal columns.
Given that the matrix is nxn, I got that the determinant is $2^n\times4$. However, this is clearly not correct since the determinant of a $4\times4$ Hadamard Matrix is 16, but according to my answer it is 64.
This is how I derived my answer:
I got $\det(-2H^2)$ by noting that the determinant of block matrices is $\det(AD - BC)$.
Where am I going wrong?
Your first equality, where you break apart the block matrix, isn't true.
A more straightforward way to find the determinant of an $n$ dimensional Hadamard matrix $H$ is to look at it this way:
Let the determinant of $H$ = $d$. If we divide a row by $\sqrt{n}$, the resulting matrix has determinant $\frac{d}{\sqrt{n}}$. Divide all rows by $n$, and the resulting matrix $H'$ is orthogonal (orthonormal columns) and thus has determinant $\frac{d}{\sqrt{n}^n} = \pm 1$ Thus, the determinant of $H$ is $\pm \sqrt{n}^n = \pm n^{n/2}$.