We calculate at the center $x$ of a complex coordinates with $g_{i\bar j} = \delta_{ij}$. Let $\omega_i = \frac{\sqrt{-1}}{2} dz^i \wedge d\bar z^i$. So we have $\omega = \sum_{i=1}^n \omega_i$. Write
$$\phi = \phi _{\bar j} \; d\bar z^j ,$$
then
$$ \partial \phi = \partial_i \phi_{\bar j} \; dz^i \wedge d\bar z^j,\ \ \bar\partial \bar\phi = \overline{\partial_{i}\phi_{\bar j}}\; d\bar z^i \wedge dz^j,$$
which gives
$$\left( \frac{\sqrt{-1}}{2}\right)^2 \partial \phi \wedge \bar\partial \bar \phi = \left( \frac{\sqrt{-1}}{2}\right)^2 \partial_i \phi_{\bar j} \overline{\partial_{l} \phi_{\bar k}} dz^i \wedge d\bar z^j \wedge d\bar z^l \wedge dz^k.$$
The above summation contains the following two types (and more):
$i=j$, $k=l$:
$$ \left(\frac{\sqrt{-1}}{2}\right)^2 \partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} dz^i \wedge d\bar z^i \wedge d\bar z^k \wedge dz^k = -\partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} \omega_i \wedge \omega_k,$$
and
$i = l$, $k=j$:
$$\left(\frac{\sqrt{-1}}{2}\right)^2 \partial_i \phi_{\bar k} \overline{\partial_{i} \phi_{\bar k} }dz^i \wedge d\bar z^k \wedge d\bar z^i \wedge dz^k = |\partial_i \phi_{\bar k}|^2 \omega_i \wedge \omega_k.$$
We care only these two types, since when $\{ i, k\} \neq \{ j, l\}$ or $i=j=k=l$ we have
$$ \left( \frac{\sqrt{-1}}{2}\right)^2 \partial_i \phi_j \overline{\partial_{l} \phi_{\bar k}} dz^i \wedge d\bar z^j \wedge d\bar z^l \wedge dz^k\wedge \omega^{n-2} = 0.$$
Hence we have
\begin{align}
\left( \frac{\sqrt{-1}}{2}\right)^2 \partial \phi \wedge \bar{\partial} \bar{\phi}\wedge \omega^{n-2} = |\partial_i \phi_{\bar k}|^2 \omega_i \wedge \omega_k \omega^{n-2}- \partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} \omega_i \wedge \omega_k \wedge \omega^{n-2}.
\end{align}
The remaining is combinatorics: since $\omega_i \wedge \omega _j = \omega _j \wedge\omega_i$, $\omega_i \wedge \omega_i = 0$,
\begin{align}
\omega^{n-2} &= ( \omega_1 + \cdots + \omega_n)^{n-2} \\
&= \sum_{i_p \neq i_q} \omega_{i_1} \wedge \omega_{i_2} \wedge \cdots \wedge \omega_{i_{n-2}} \\
&= (n-2)! \sum_{i\neq k} \omega_1 \wedge \cdots \wedge \widehat{\omega_i}\wedge \cdots \wedge\widehat{\omega_k}\wedge \cdots \wedge \omega_n,
\end{align}
here $\widehat{\omega_i}$ means $\omega_i$ is excluded. The last equality follows from the fact that there are $(n-2)!$ ways to form $\omega_1 \wedge \cdots \wedge \widehat{\omega_i}\wedge \cdots \wedge\widehat{\omega_k}\wedge \cdots \wedge \omega_n$.
Thus
\begin{align}
\left( \frac{\sqrt{-1}}{2}\right)^2 \partial \phi \wedge \bar{\partial} \bar{\phi}\wedge \omega^{n-2} &= |\partial_i \phi_{\bar k}|^2 \omega_i \wedge \omega_k \omega^{n-2}- \partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} \omega_i \wedge \omega_k \wedge \omega^{n-2}. \\
&=(n-2)!\left( \sum_{i,k} |\partial_i \phi_{\bar k}|^2 - \sum_{i,k}\partial_i \phi_{\bar i} \overline{\partial_{k} \phi_{\bar k}} \right) \omega_1\wedge\cdots \wedge \omega^n\\
&= \frac{1}{n(n-1)} (|\partial \phi|^2 - |\bar\partial^* \phi|^2 ) \omega^n
\end{align}
Since
$$\omega^n = n!\; \omega_1\wedge \cdots\wedge \omega_n,$$
$$ |\partial \phi|^2 = \sum_{i,k} |\partial_i \phi_{\bar k}|^2$$
and (see here)
$$\bar\partial^* \phi = -\sum_i \partial_i \phi_{\bar i}$$
at $x$.
Best Answer
On any Kahler manifold, the Kahler identities say that $[\Lambda, \bar\partial] = -i \partial^\dagger$, where $\Lambda$ is the adjoint of the Lefchetz operator and $\partial^\dagger$ is the adjoint of $\partial$. For a smooth function $f$, we know that $\partial^\dagger f = 0$ for degree reasons. Thus $$ \Delta_\partial f = (\partial \partial^\dagger + \partial^\dagger \partial)f = \partial^\dagger \partial f = i(\Lambda \bar\partial - \bar\partial \Lambda) \partial f = i \Lambda \bar\partial\partial f $$ as $\Lambda \partial f = 0$ for degree reasons. As $\Lambda$ is complex linear and $\Delta_d = 2\Delta_\partial$, we conclude that $$ \Delta_d f = 2i \Lambda \bar\partial\partial f = \frac 2i \Lambda \partial \bar\partial f $$ Taking the product with the volume form $\omega^n/n!$ we get $$ \Delta_d f \, \omega^n/n! = \Lambda \frac 2i \partial\bar\partial f \, \omega^n/n! = \frac 2i \, \partial\bar\partial f \wedge \omega^{n-1}/(n-1)!. $$ On a surface, where $n = 2$, this agrees with what your calculations gave.