I'm familiar with the usual proof $\det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A)$ which only works in fields of characteristic not equal to 2.
To get a proof that works in characteristic 2 I can bash out the determinant using the Leibniz expansion,
$$\det A = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n A_{i, \sigma(i)}$$
Since
$$\prod_{i=1}^n A_{i, \sigma^{-1}(i)} = \prod_{i=1}^n A_{\sigma(i), i} = (-1)^n \prod_{i=1}^n A_{i, \sigma(i)}$$
and since a permutation has the same sign as its inverse, for odd $n$ each permutation cancels with its inverse in the Leibniz expansion, except for permutations which are their own inverses. But such permutations are the product of disjoint transpositions and since $n$ is odd they must therefore leave at most one element fixed. Since $A$ is skew-symmetric, on-diagonal terms are zero and those permutations contribute nothing.
I was wondering whether there is a simpler/nicer proof that works in all fields?
@ Brain, $\det(A)=P((a_{i,j})_{i,j})$ where $P$ is a polynomial. This polynomial is formally $0$. Then your equality is valid in any field and, even, in any commutative ring.