= Solution
Let $g$ be the <Riemannian metric> underlying the <Hermitian metric> and $J$ its compatible <complex structure>. The <fundamental form of a Hermitian manifold> is $\omega(u,v)=g(Ju,v)$. It is a real alternating form of type $(1,1)$. At a point choose an orthonormal real coframe $x^1,y^1,\ldots,x^n,y^n$ adapted to $J$, with $x^j+i y^j$ of type $(1,0)$. Then
$$
\omega=\sum_{j=1}^n x^j\wedge y^j,
\qquad
\boxed{dV_g=x^1\wedge y^1\wedge\cdots\wedge x^n\wedge y^n=\frac{\omega^n}{n!}.}
$$
The orientation here is the complex orientation. In coordinates for which $g=\sum_j((dx^j)^2+(dy^j)^2)$, the same convention reads $\omega=(i/2)\sum_j dz^j\wedge d\bar z^j$.
Use the <complex Hodge star operator>, the complex-linear extension of the real <Hodge star operator>. With the pointwise <Hermitian inner product> linear in its first variable, its defining identity is
$$
\alpha\wedge*\bar\beta=\langle\alpha,\beta\rangle_g\,dV_g
$$
for forms of the same total degree. It maps type $(p,q)$ to $(n-q,n-p)$ and satisfies $*^2=(-1)^{p+q}$ in real dimension $2n$. It commutes with <complex conjugation>. These conventions distinguish it from the <conjugate-linear Hodge star> $\alpha\mapsto*\bar\alpha$.
Write a top holomorphic-type form as $\eta=a\,\zeta^1\wedge\cdots\wedge\zeta^n$, where $\zeta^j=x^j+i y^j$. On each oriented two-plane, $*x^j=y^j$ and $*y^j=-x^j$, so the two-dimensional star of $\zeta^j$ is $-i\zeta^j$. In taking the star of their product, regrouping complementary one-form factors across the $n$ two-planes contributes $(-1)^{n(n-1)/2}$. Thus the <Hodge star on top holomorphic forms> is
$$
*\eta=(-1)^{n(n-1)/2}(-i)^n\eta,
\qquad
\boxed{*\eta=(-1)^{n(n+1)/2}i^n\eta.}
$$
This is pointwise and requires neither holomorphicity of the coefficient $a$ nor closedness of $\omega$.
To compute the <formal adjoint>, take compactly supported forms $\alpha$ of degree $k-1$ and $\beta$ of degree $k$. The <Stokes theorem> gives
$$
\int_X d\alpha\wedge*\bar\beta
=(-1)^k\int_X\alpha\wedge d(*\bar\beta).
$$
Since the degree of $d(*\bar\beta)$ is $2n-k+1$, the identity for $*^2$ makes the right side equal to $\int_X\alpha\wedge*\overline{(-*d*\beta)}$. Consequently the <formal adjoint> of $d$ in even real dimension is $d^*=-*d*$. Splitting by type, $-*\partial*$ lowers antiholomorphic degree, while $-*\bar\partial*$ lowers holomorphic degree. Orthogonality of distinct types therefore yields
$$
\boxed{\bar\partial^*=-*\partial*,\qquad
\partial^*=-*\bar\partial*.}
$$
The argument applies locally with compact support, and globally without support restrictions on a compact manifold without boundary.
The <Dolbeault Laplacian> is $\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial$. A form is $\bar\partial$-harmonic when it lies in its kernel. On a compact <Hermitian manifold>, the adjoint identities imply
$$
\langle\Delta_{\bar\partial}\alpha,\alpha\rangle_{L^2}
=\|\bar\partial\alpha\|_{L^2}^2+\|\bar\partial^*\alpha\|_{L^2}^2.
$$
Thus harmonicity is equivalent to both first-order equations vanishing. The <Dolbeault Hodge decomposition on a compact Hermitian manifold> states that the harmonic space $\mathcal H^{p,q}_{\bar\partial}$ is finite-dimensional and that smooth forms have the orthogonal direct-sum decomposition
$$
\Omega^{p,q}(X)=\mathcal H^{p,q}_{\bar\partial}
\oplus\bar\partial\Omega^{p,q-1}(X)
\oplus\bar\partial^*\Omega^{p,q+1}(X).
$$
To deduce the identification with <Dolbeault cohomology>, let $\bar\partial\alpha=0$ and decompose $\alpha=h+\bar\partial u+\bar\partial^*v$. The closedness of $\alpha$ makes it orthogonal to the entire image of $\bar\partial^*$. Taking its inner product with $\bar\partial^*v$ and using the orthogonality of the decomposition gives $\|\bar\partial^*v\|^2=0$. Hence $\alpha=h+\bar\partial u$, so each cohomology class has a harmonic representative. If a harmonic form equals $\bar\partial u$, its squared norm is $\langle\bar\partial^*h,u\rangle=0$, so the representative is unique. We have proved
$$
\boxed{\mathcal H^{p,q}_{\bar\partial}\longrightarrow H^{p,q}_{\bar\partial}(X),\quad h\longmapsto[h],\text{ is an isomorphism}.}
$$
For the complementary types, define $C\alpha=*\bar\alpha$. It maps $(p,q)$ to $(n-p,n-q)$ and is conjugate-linear. If $\alpha$ is harmonic, then $\bar\partial\alpha=0$ and $\partial(*\alpha)=0$ by the adjoint formula. Conjugating the latter equation gives $\bar\partial(*\bar\alpha)=0$. Also, with $k=p+q$,
$$
\bar\partial^*(C\alpha)
=-*\partial*(*\bar\alpha)
=-(-1)^k*\partial\bar\alpha=0,
$$
since $\partial\bar\alpha=\overline{\bar\partial\alpha}$. Therefore $C$ carries harmonic forms to harmonic forms, and $C^2=(-1)^k$ makes it a bijection. This proves <complementary Dolbeault harmonic types on a Hermitian manifold>:
$$
\boxed{\dim_{\mathbb C}\mathcal H^{p,q}_{\bar\partial}
=\dim_{\mathbb C}\mathcal H^{n-p,n-q}_{\bar\partial}.}
$$
The metric supplies a natural conjugate-linear isomorphism. If a complex-linear isomorphism is desired, choose a basis $h_j$ of the first space and send it to the basis $Ch_j$ of the second, extending linearly. This last choice is not canonical; no <Kähler> assumption is needed for either the dimension equality or the harmonic bijection.
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