To make notation easier I will write $z, x, y$ instead of $dz, dx, dy.$ Furthermore, this all boils down to linear algebra, so I will work with a real vector space $V$ with an inner product and compatible complex structure $J$ where $y_i = J(x_i)$. Further assume that the $x_i$'s and $y_i$'s are orthogonal. Remember how the inner product on $V$ extends to $\bigwedge^k V$: we declare that monomials in the $x$ and $y$'s of length $k$ are orthonormal.
Set $z_j = x_j + i y_j$.
I will call a $(p,q)$ form a monomial if it is of the form $z_I \wedge \bar{z}_J$. I will often omit wedge product symbols and write $z_I \bar z_J$.
I will record some observations:
Lemma $1$. $\bar\ast$ takes complex $k$-forms into $(n-k)$-forms.
Proof This is the (conjugate) of the complexification of the real Hodge star, which has this same property.
Lemma $2$. $\bar\ast(z_I\bar{z}_J) = b z_{I^c} \bar{z}_{J^c}$ where $I^c$ is the complementary set of indices, and $b$ is some nonzero constant.
Proof. Say $z_I \bar{z}_J$ is a $(p,q)$-form. Write $\bar\ast (z_I \bar{z}_J) = \sum_{K,L} b_{K,L} z_K\bar{z}_L$, where $|K| + |L| = 2n-(p+q)$ (this implicitly uses lemma $1$ to get the sizes of the multi-indices right). Since $z_I\bar{z}_J \wedge \bar\ast(z_I\bar{z}_J) = ||z_I \bar{z}_J||^2 \cdot \text{vol}$, we note that $\sum_{K,L} b_{K,L} z_I\bar{z}_J z_K\bar{z}_L$ must equal a positive multiple of the volume form. But $z_I \bar{z}_J z_K\bar{z}_L=0$ unless $K= I^c$ and $L=J^c$. You can think about why this last step is true; the reasoning should morally be that there will either be more than $n$ $z$'s or more than $n$ $\bar{z}$'s, killing the product $z_I \bar{z}_Jz_K\bar{z}_L$. The constant $b := b_{I^c, J^c}$ is nonzero because we need to have
$$
b z_I\bar{z}_J z_{I^c}\bar{z}_{J^c} = ||z_I\bar{z}_J||^2 \cdot \text{vol} \neq 0.
$$
Computing $\bar\ast\alpha$ for $\alpha = z_2\bar{z}_1\bar{z}_4$, by lemma $2$ we already know that
$$
\bar\ast\alpha = b \cdot z_1 z_3 z_4 \bar{z}_2\bar{z}_3,
$$
and it remains to only calculate the value of $b$. We'll need $||\alpha||$ for this.
Expand $\alpha$ into $x$ and $y$'s:
$$\alpha = z_2\bar{z}_1\bar{z}_4 = x_2x_1x_4 -i x_2y_1x_4 + i y_2x_1x_4 + y_2y_1y_4 -ix_2x_1y_4 - x_2y_1y_4 + y_2x_1y_4 - iy_2y_1y_4.$$
Since these monomials in $x$ and $y$ are all orthonormal and distinct, we can see that $||\alpha||^2 = 8$.
So,
$$
z_2 \bar{z}_1 \bar{z}_4 \wedge b z_1 z_3 z_4\bar{z}_2\bar{z}_3 = 8 \cdot \text{vol}.
$$
Next, you can check that
$$
z_2 \bar{z}_1 \bar{z}_4 z_1 z_3 z_4\bar{z}_2\bar{z}_3 = - z_1\bar{z}_1z_2\bar{z}_2 z_3\bar{z}_3z_4\bar{z}_4 = -\left(\frac{2}{i}\right)^4 \cdot \text{vol}
$$
and so we find that
$$
-b \left(\frac{2}{i}\right)^4 = 8,
$$
or $b=-\frac{1}{2}$. Therefore
$$
\bar\ast(z_2\bar{z}_1\bar{z}_4) = -\frac{1}{2} z_1z_3z_4\bar{z}_2\bar{z}_3.
$$
Your second question: no. If $\alpha = z_I \bar{z}_J$, then $\bar{\alpha} = \bar{z}_I z_J \neq z_J \bar{z}_I$ in general. You would have to figure out the number of permutations needed and count the right number of signs.
I implicitly answered your last question above, but I'll state it clearly here: you permute $dz$ and $d\bar{z}$'s like ordinary $1$-forms, so
$$
dz_1 \wedge d\bar{z}_3 \wedge d z_2 = - dz_1 \wedge dz_2 \wedge d\bar{z}_3.
$$
I now figured out something, but I am not sure whether my calculations are correct, so please tell me if my approach is acceptable.
Let $\alpha \in \Lambda^p$ such that $\alpha = \frac{1}{p!} \; \alpha_{k_1 \ldots k_p} \; dx^{k_1} \wedge \ldots dx^{k_p}$ and let $\beta = dx^{j_1} \wedge \ldots \wedge dx^{j_p}$. Then by assumption
\begin{align}
\alpha \wedge \star \beta \enspace &= \enspace \frac{\sqrt{\det g}}{(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_n} \; \alpha \wedge\Big(dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \Big) \\
&= \enspace \frac{\sqrt{\det g}}{p! \, (n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_n} \; \alpha_{k_1 \ldots k_p} \; \Big(dx^{k_1} \wedge \ldots \wedge dx^{k_p} \Big) \wedge \Big(dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \Big)
\end{align}
By renaming the indices, one obtains
\begin{align}
\alpha \wedge \star \beta \enspace &= \enspace \frac{\sqrt{\det g}}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \alpha_{k_1 \ldots k_p} \; \underbrace{\Big(dx^{k_1} \wedge \ldots \wedge dx^{k_p} \Big) \wedge\Big(dx^{k_{p+1}} \wedge \ldots \wedge dx^{k_n} \Big)}_{= \; \varepsilon^{k_1 \ldots k_n} \; dx^1 \wedge \ldots \wedge dx^n} \\
&= \enspace \frac{\sqrt{\det g}}{p! \, (n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \alpha_{k_1 \ldots k_p} \; \; \varepsilon^{k_1 \ldots k_n} \; \Big( dx^1 \wedge \ldots \wedge dx^n \Big)
\end{align}
Using $dV =\sqrt{\det g} \cdot dx^1 \wedge \ldots \wedge dx^n$, one gets
\begin{align}
\alpha \wedge \star \beta \enspace &= \enspace \frac{dV}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \varepsilon_{j_1 \ldots j_p k_{p+1} \ldots k_n} \; \varepsilon^{k_1 \ldots k_n} \; \alpha_{k_1 \ldots k_p}\\
&= \enspace \frac{dV}{p! \,(n-p)!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; (n-p)! \; \delta^{k_1 \ldots k_p}_{j_1 \ldots j_p} \; \alpha_{k_1 \ldots k_p} \\
&= \enspace \frac{dV}{p!} \; g^{i_1 j_1} \cdots g^{i_p j_p} \; p! \; \alpha_{[j_1 \ldots j_p]} \\
&= \enspace dV \; g^{i_1 j_1} \cdots g^{i_p j_p} \; \alpha_{j_1 \ldots j_p}
\end{align}
which coincides with the inner product. Here, the antiymmetry of $\alpha$ was used. Note that the factor of $\tfrac{1}{p!}$ within the inner product arises from a $p$-form $\gamma = \tfrac{1}{p!} \, \gamma_{m_1 \ldots m_p} \, dx^{m_1} \wedge \ldots \wedge dx^{m_p}$ being inserted into the inner product, i.e. $(\alpha, \gamma)$.
Please tell me what you think.
Best Answer
There are a few conventions here. You can check whether it is linear as follows: $$\forall\alpha:\alpha\wedge \overline{\ast(k\beta)}=\langle \alpha,k\beta\rangle (\ast1)=\bar k\langle \alpha,\beta\rangle (*1)=\alpha\wedge\overline{k\ast(\beta)}\Longrightarrow \ast(k\beta)= k\ast(\beta)$$
On the other hand, it is also common in the literature to define the Hodge star operator by the equation $\alpha\wedge\ast\beta=\langle \alpha,\beta\rangle(\ast1)$. Likewise, we can check $$\forall\alpha:\alpha\wedge\ast(k\beta)=\langle\alpha,k\beta\rangle(\ast1)=\bar k\langle\alpha,\beta\rangle(\ast 1)=\alpha\wedge\bar k\ast(\beta)\Longrightarrow\ast(k\beta)=\bar k\ast(\beta)$$
This becomes antilinear. Note that the first one sends a $(p,q)$-form to $(n-q,n-p)$ form meanwhile the second definition sends a $(p,q)$-form to $(n-p,n-q)$ form.