In real case, $$ \ast (e_1\cdots e_k)=e_{k+1}\cdots e_n$$ on $\mathbb{R}^n$, where $e_i$ is $1$-form and $e_1\cdots e_n$ is volume on $\mathbb{R}^n$
We will extend to complex case. Define $$ dz_k:= e_k + ie_{n+k} $$ On $\mathbb{C}^n=\mathbb{R}^{2n}$ we have a volume form : $$ dv:= e_1e_{n+1}\cdots e_n e_{2n} $$
Hence we have for instance : $$
\ast (dz_1)=e_{n+1}e_2\cdots e_{2n} - i e_1e_2\cdots e_{2n} $$
That is, $\ast$ is $\mathbb{C}$-linear.
Problem : We want to prove that $$ \ast : \Lambda^{p,q}\rightarrow \Lambda^{n-q,n-p} $$ where $\Lambda^{p,q}$ is a space of elements of $$\alpha :=dz_{i_1}\cdots dz_{i_p}d\overline{z}_{j_1}\cdots d\overline{z}_{j_q} $$
Proof : (1) Note that $$ \frac{i}{2} dz_1d\overline{z}_1 = e_1e_{n+1} $$ so that $$ dv = (\frac{i}{2})^n dv_0,\ dv_0:= dz_1d\overline{z}_1 \cdots dz_nd\overline{z}_n $$
Hence we have a map $f$ s.t. $$ A\in \mathbb{C},\ f (A\alpha )=A f(\alpha),\ \alpha f(\alpha)=dv_0 $$
(2) If $n=1$, $$ \ast dz_1=-idz_1 $$
Hence problem holds.
(3) If $n=2,\ a=dz_2$ then by direct computation $$ \ast a = C_1 dz_1 d\overline{z}_1 dz_2 = C_2\overline{ f(a)} $$ for some $C_i\in \mathbb{C}$.
But general case, I have no idea. Thank you for your attention.
The definition of the Hodge star operator depends on the definition of the $L^2$ inner product, which depends on the definition of the metric on the exterior algebra $\bigwedge^{\bullet} (T^*M)$ on a compact oriented real manifold $M$.
Take a section $\chi_1,\dots,\chi_n$ of $T^*M$, then for each $\mathbb{Z}\ni k>1$, the $k$-forma $\chi_{i_1}\wedge\cdots\chi_{i_k}$, where $1\leq i_1<\cdot<i_n\leq n$ are strictly increasing sequences of $k$-tuples in $[n]:=\{1,\dots,n\}$ form a basis for $\bigwedge^k(T^* M)$.
Then, if we require that basis to be orthonormal if $\chi_1,\dots,\chi_n$ forms an orthonormal frame on $T^*M$. We can extend by bilinearity to a non-degenerate inner product on $\bigwedge^k(T^*M)$. One can show (using the Cauchy-Binet formula), that if $\alpha_1,\dots,\alpha_k,\beta_1,\dots,\beta_k$ are in $T_p^*M$, then
$$\langle \alpha_1\wedge\dots\wedge \alpha_k,\beta_1\wedge\dots\beta_k\rangle = \det((\langle\alpha_i,\beta_j\rangle)_{i,j\in [k]})$$
Since, $M$ is oriented if and only if there exists a unique $n$-form $\Omega$ that is $1$ on every positively oriented orthonormal basis of $T_pM$.
We define $*$ to be such that, for $k$-forms $\alpha,$ and $\beta$
$$\alpha\wedge *\beta = \langle \alpha,\beta\rangle\Omega$$
Now, since the manifold $M$ is compact, the integral $$(\alpha,\beta):= \int_M \alpha\wedge *\beta = \int_M \langle\alpha,\beta\rangle \Omega$$ is a positive definite inner product on $k$-forms.
Now, here's the crucial observation, when we extend this inner product from real forms to complex forms, we need to add complex conjugation to the second factor to keep it positive definite. So, on $A^k(M,\mathbb{C})$ the space of sections of the $k^{\rm th}$ exterior power of the cotangent bundle with complex coefficients, the $L^2$ inner product is, $$(\alpha,\beta):= \int_M \alpha\wedge \overline{*\beta}$$
Now, we need $\overline{*\beta}$ to be a $(n-p,n-q)$-form for $\alpha\wedge \overline{*\beta}$ to be a $2n-$form which is the desired real dimension of the manifold $M$, such that $M$ has complex dimension $n$.
Since, complex conjugation sends $(i,j)-$forms to $(j,i)-$forms. $*\beta$ has to be an $n-q,n-p$ form!