A complex $n$-dimensional vector space $V$ can be thought of as a real $2n$-dimensional vector space equipped with a map $J:V \to V$ with $J^2 = -I$. Complex-linear maps are then linear maps $V \to V$ which commute with $J$. One can think of $J$ as an infinitesimal rotation, so that $\exp(tJ)$ gives a family of rotations of this space, and $\mathbb R$-linear maps $V \to V$ are complex-linear if they respect this family.
From this point of view, or some other geometric point of view, is there a nice interpretation of the complex determinant $\det_{\mathbb C} L$ of a complex-linear map $L: V \to V$? Or, almost the same question, is there a geometric interpretation of the unique (up to scaling by complex numbers) antisymmetric complex-multilinear $n$-form $\operatorname{vol}_{\mathbb C}: V \times V \times ... \times V \to \mathbb C$?
The norm is fairly easy to interpret. $| \det_{\mathbb C} L |^2 = |\det_{\mathbb R} L|$. One way to see this is to look at the diagonalization of $L$ over $\mathbb C$. This also gives you a way to interpret the argument, as the total amount of rotation in all the invariant subspaces of $L$.
Is there a geometric interpretation of $\det_{\mathbb C} L$, not just its norm, which does not require one to diagonalize the matrix first?
Even the special case when $L$ is unitary is of interest.
Note to moderators: I tried to edit my deleted answer, but then people would have to vote to undelete my post. So I have decided to add another answer than my deleted one. Besides, it is completely different from what I had written about 3 years ago!
Let $V = \mathbb{C}^n$, $h(-,-)$ denote the standard hermitian inner product on $V$ and denote by $\Omega \in \Lambda^n V^*$, normalized so that $$ \Omega(e_1, \ldots, e_n) = 1. $$
I claim that this data allows us to define what I would like to call the hermitian cross product of $n - 1$ vectors in $V$. Indeed, let $v_1, \ldots, v_{n-1} \in V$. Denote by $*_h(v_1 \wedge \cdots \wedge v_{n-1}) \in V$ the unique vector, say $w \in V$ for short, such that
$$ \Omega(v_1, \ldots, v_{n-1}, v) = h(v, w) $$ for all $v \in V$ (my convention is that $h(-,-)$ is antilinear in the second argument).
Taking $v = w$, we get
$$ \Omega(v_1, \ldots, v_{n-1}, w) = h(w, w) = \lVert w \rVert^2 \geq 0.$$
Thus $w$ is a vector in $V$ which is orthogonal to the span of $v_1, \ldots, v_{n-1}$ and such that the above complex determinant is actually not only real, but also non-negative. We will call $w$ the hermitian cross product of $v_1, \ldots, v_{n-1}$ (there is probably another name for it in the literature, I am sure).
In terms of coordinates, $w = (w_1, \ldots, w_n)$, it is easy to show that $$ w_i = (-1)^{i+n} \overline{m_{in}(A)} $$
where $m_{ij}(A)$ is the $ij$-th minor of $A$ and $A$ is the complex $n$ by $n-1$ matrix whose $j$-th column is $v_j$, for $j = 1, \ldots, n-1$.
If we take $v = v_n$ we then get $$ \Omega(v_1, \ldots, v_n) = h(v_n, w). $$
So we can view the complex determinant of $v_1, \ldots, v_n$ as the hermitian inner product of, say $v_n$, with a specific vector $w$, which is nothing but the hermitian cross product of the first $n - 1$ vectors ($v_1, \ldots, v_{n-1}$).
Our task is now much easier! Thinking about the real picture, orthogonaly project $v_n$, now thought of as a point in $\mathbb{R}^{2n}$, orthogonally onto the real span of $w$ and $Jw$. Note that we can identify the real span of $w$ and $Jw$ with $\mathbb{C}$, by mapping $aw + bJw$ onto $\lVert w \rVert (a + bi)$.
I believe that the complex determinant of $v_1, \ldots, v_n$ is actually $\lVert w \rVert$ times the orthogonal projection of $v_n$ onto the real span of $w$ and $Jw$, itself identified with $\mathbb{C}$. This in particular explains the real and imaginary part of the complex determinant and thus, in particular, its phase.
Edit 1: Here is another approach towards understanding the phase of the complex determinant, in the case of $2$ by $2$ complex matrices, using the Hopf map.
The Hopf map $h: S^3 \to S^2$ has the property that $h(e^{i\theta} \psi) = h(\psi)$, for any $\psi \in S^3 \subset \mathbb{C}^2$. So we can think of a point on $S^3$ simply as a triple consisting of $(p, v, \nu)$, where $p \in S^2$, $v$ is a unit vector in $\mathbb{R}^3$ which is orthogonal to $p$ in $\mathbb{R}^3$ but which is based at $p$. So we can think of $v \in T_p(S^2)$ as a unit vector living in the tangent plane to $S^2$ at $p$. Finally, $\nu$ is a point in the non-trivial double cover of $SO(3)$ lying above $(p, v, p \times v) \in SO(3)$.
Given $\psi_a \in S^3$ for $a = 1, 2$, which we assume to be linearly independent in $\mathbb{C}^2$ (over $\mathbb{C}$) WLOG, we then form the triples $(p_a, v_a, \nu_a)$ for $a = 1, 2$, corresponding to $\psi_a$ for $a = 1, 2$, respectively.
In this setup, how can we recover the phase of the complex determinant of the $2$ by $2$ matrix $(\psi_1, \psi_2)$?