Davide's answer tells most of the story, in particular giving the proof for the determinant, but not quite all of it, so I want to supplement it with a couple of remarks.
I think that it is more common to replace Davide's matrix $B$ with a real matrix. This can be achieved by conjugating it with matrix of the block form $D=\pmatrix{I&0\cr0&iI\cr},$ when $D^{-1}BD=\pmatrix{A_1&-A_2\cr A_2 &A_1\cr}.$ Because conjugation preserves the determinant, Davide's calculation tells that here we also have $\det(D^{-1}BD)=\det(B)=|\det A|^2.$ Further conjugating (shuffling rows and columns) allows us to replace each and every complex entry $z=a+bi$ with a $2\times2$ block $ (z)=\pmatrix{a&-b\cr b&a\cr}. $ Doing it this way makes it clear that if $A$ represents a linear mapping $T$ from $V=\mathbf{C}^n$ to itself with respect to basis $v_1,v_2,\ldots,v_n$, then $D^{-1}BD$ represents the same mapping $T$, when we view $V$ as a real vector space of dimension $2n$ and use the basis $v_1,v_2,\ldots,v_n,iv_1,iv_2,\ldots,iv_n.$
The extra shuffling I talked about would reorder this latter basis to $v_1,iv_1,\ldots$.
A geometric interpretation of this is that $\det B$ gives the scaling of Lebesgue measure (or hypervolumes) of a box $K=\prod_{i=1}^{2n}[c_i,d_i]$ of real dimension $2n$ under $T$: $ \det (B)=\frac{m(T(K))}{m(K)}.$
$\det A$ does the same, but because the coordinates are complex there, we need to use $|\det A|^2$ to get the scaling right. This is seen already in the complex plane, where multiplication by $a+bi$ multiplies the areas of rectangles by a factor of $a^2+b^2$.