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 $x$.
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.
Best Answer
Let $A = A_{\bar{i}} \overline{dz^{i}}$ be any $(0,1)$-form (we want to use $A = \overline\partial f$ later). Using the definition of $\overline\partial ^*$: for all test function $\varphi$, write $A^i = g^{\bar j i} A_{\bar j}$, \begin{align*} \int_M \varphi\overline{\overline\partial ^* A} dV &=\int_M (\overline\partial \varphi)_{\bar i} \overline{A^{i}}dV \\ &=\int_M \frac{\partial \varphi}{\partial \bar{z^i}}\overline{A^{i}} (\sqrt{-1})^n G dz^N \wedge \overline{dz^N}, \end{align*}
Where $dz^N = dz^1 \wedge \cdots \wedge dz^n$ and $G = \det (g_{i\bar j})$. Integration by part gives
\begin{align*} \int_M \varphi\overline{\overline\partial ^* A} dV &= - \int_M \varphi \partial_{\bar i}(\overline{A^{i}} G) (\sqrt{-1})^n dz^N \wedge \overline{dz^N}\\ &= -\int_M \varphi \left(\overline{\partial_i A^i + A^i G^{-1} \partial_i G} \right) (\sqrt{-1})^n G dz^N \wedge \overline{dz^N} \\ &= -\int_M \varphi \left(\overline{\partial_i A^i + A^i \partial_i(\log G)} \right)dV. \end{align*}
Thus \begin{equation}\tag{1} \overline\partial ^* A= - (\partial_i A^i+ (\partial_i \log G) A^i), \end{equation} Now calculate: \begin{align*} \partial_i A^i + \partial_i \log G A^i &= \partial_i (g^{\bar j i} A_{\bar j}) + g^{ m \bar k} \frac{\partial g_{m\bar k}}{\partial z^i} g^{\bar j i} A_{\bar j}\\ &= g^{\bar j i} \frac{\partial A_{\bar j}}{\partial z^i} + \left(\frac{\partial g_{\bar j i}}{\partial z^i} + g^{m \bar k} \frac{\partial g_{m\bar k}}{\partial z^i} g^{\bar j i} \right)A_{\bar j} \end{align*}
Now we use the Kähler condition: in particular, we have $$ \frac{\partial g_{m\bar k}}{\partial z^i} = \frac{\partial g_{i\bar k}}{\partial z^m}, $$ thus \begin{align*} g^{m \bar k} \frac{\partial g_{m\bar k}}{\partial z^i} g^{\bar j i} &= g^{m \bar k} \frac{\partial g_{i\bar k}}{\partial z^m} g^{\bar j i} \\ &= - g^{m \bar k} \frac{\partial g^{\bar j i}}{\partial z^m} g_{i\bar k} \\ &=-\frac{\partial g^{\bar j i}}{\partial z^i} \end{align*}
Then we have $$\overline\partial ^* A = - g^{\bar j i} \frac{\partial A_{\bar j}}{\partial z^i}$$
and setting $A = \overline\partial f$ gives
$$ \Delta f = 2 \overline\partial^* \overline\partial f = -2 g^{\bar j i} \frac{\partial ^2 f}{\partial z^i \partial \bar z^j}.$$