Consider a compact Kaehler manifold $X$ of dimension $d=3$. The K"ahler form in local coordinates $z^i,\bar{z}^i$ is $$ \omega = \omega_{i \bar{j}}dz^i \wedge d\bar{z}^j, $$ with $\omega_{i \bar{j}} = \frac{i}{2} g_{i \bar{j}}$, where $g_{i \bar{j}}$ is Hermitian.
I want to explicitly show that the volume given by $\int_X \omega^n $ yields $$ A \int_X \sqrt{\mathrm{det}(g_{i\bar{j}}) } dz^1\wedge dz^2 \wedge dz^3 \wedge d\bar{z}^1 \wedge d\bar{z}^2 \wedge d\bar{z}^3, $$ where $A$ is some overall constant.
The closest I arrive at my intended result is: $$ \int_X \epsilon^{i k m} \epsilon^{\bar{j} \bar{l} \bar{n} } ~ \omega_{i \bar{j}} \omega_{k \bar{l}} \omega_{m \bar{n} } ~ dz^1\wedge dz^2 \wedge dz^3 \wedge d\bar{z}^1 \wedge d\bar{z}^2 \wedge d\bar{z}^3, $$ using the antisymmetry of the wedge product and the identity $$ dz^i \wedge dz^k \wedge dz^m = \epsilon^{i k m} dz^1\wedge dz^2 \wedge dz^3 $$
and its complex conjugate.
I'm almost convinced that I can identify $ \epsilon^{i k m} \epsilon^{\bar{j} \bar{l} \bar{n} } ~ \omega_{i \bar{j}}\omega_{k \bar{l}} \omega_{m \bar{n} } = \mathrm{det} (\omega_{i \bar{j}})$, but not 100% sure.
(FWIW, I have seen a version of this where the $dz^i , d\bar{z}^j$'s are turned into real forms $dx^i, dy^j$, but I want to avoid going down that route for the moment and express the result in complex coordinates.)
EDIT: The factor $\sqrt{\mathrm{det}(g_{i\bar{j}}) }$ should not have a square root if it is the hermitian metric, however the real Riemannian metric it is associated to will be square rooted, i.e. $$ \sqrt{\mathrm{det}(g_{\mu \nu}(x,y) )} = \mathrm{det}(g_{i\bar{j}}(z, \bar{z}) ) $$