Given a skew-hermitian matrix $A \in \mathbb{C}^{N\times N}$, then $A = -A^H = -(A^*)^{T}.$ We can also say that $A^T = (-(A^*)^T)^T = -A^*.$ Thus, when computing the determinant we get $$ \det(A) = \det(A^T) = \det(-A^*) = (-1)^N\det(A^*) = (-1)^N(\det(A))^*. $$ For $N$ even, we have $\det(A) = (\det(A))^*$, and we can conclude that $\det(A)$ is real.
For $N$ odd, we have $\det(A) = -(\det(A))^*$.
Is the only conclusion that $\det(A) = 0$ for odd $N$?
Let $y = det(A)$. For $N$ odd, we have $y = -\bar y$. Therefore, $y$ is imaginary (or can be $0$, too).