So, what is the precise expression of $⟨v,w⟩$?
Let $(M,g)$ be an orientable Riemannian manifold of dimension $n$. For any $0 \le k \le n$, there is a unique extension of the metric $g$ to the metric
$⟨ \, ,\, ⟩: \Omega^k(M) \times \Omega^k(M) \to \mathbb{R}$ such that
for any O.N.B. $ \{e^1, ...,e^n \} $ of one-forms, the basis of
$\Omega^k(M)$ is an O.N.B. with respect to $⟨ \, ,\, ⟩$
$$\text{i.e. } ⟨ e^{i_1} \wedge...\wedge e^{i_k} \, ,\, e^{j_1} \wedge...\wedge e^{j_k}⟩ = \delta^{i_1 j_1}...\delta^{i_k j_k} \, .\tag4$$
More generally, for $v^1,...,v^k,w^1,...,w^k \in \Omega(M)$,
$$ ⟨v^1\wedge...\wedge v^k,w^1\wedge...\wedge w^k⟩ := \mathrm{det}(g(v^i,w^j)) \,. \tag5$$
It is easy to check that this form is bilinear, symmetric and positive-definite.
Now suppose $ v := v_{i_1...i_k} e^{i_1} \wedge...\wedge e^{i_k}$ and $w:=w_{j_1...j_k}e^{j_1} \wedge...\wedge e^{j_k}$ are the coordinate representations of the $k$-forms $v$ and $w$. Therefore,
\begin{align*}
⟨v,w⟩ &= v_{i_1...i_k} \ w_{j_1...j_k} \,\delta^{i_1 j_1}...\delta^{i_k j_k} \\
&= v_{i_1...i_k} \ w^{i_1...i_k} \,. \tag8 \\
\end{align*}
How to derive the coordinate expression of the Hodge dual?
To begin with, I will take your definition and write it down as follows.
The Hodge dual is the unique isomorphism
\begin{align*} \star:\Omega^k(M) &\to \Omega^{n-k}(M), \\ \omega
&\mapsto \star \omega \\ \end{align*}
such that the following holds:
$$ \forall \omega, \eta \in \Omega^k(M): \omega \wedge \star \eta =
⟨\omega,\eta⟩ \ vol$$
where $vol := \sqrt{g} \ \mathrm{d} x^1
\wedge ... \wedge \mathrm{d}x^n$ is the volume form.
I have purposefully deviated from the notation used in (1) to avoid confusion between $w$ and $\omega$. Moreover, I believe that (2) is not correct at all. I will derive the correct coordinate representation below. Since the Hodge star operator is a $C^\infty (M)$-linear map, it suffices to evaluate it on the basis elements.
Proposition:
$$ \star (\mathrm{d}x^{i_1}\wedge...\wedge \mathrm{d}x^{i_k}) =
\frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1
... j_n} \ \mathrm{d}x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}\,.\tag9$$
This immediately implies
$$ \boxed{(\star \omega)_{j_{k+1} ... j_n} = \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \omega_{i_1 ... i_k}} \,. \tag{10}$$
Note the structural difference between this (10) and your claim (2).
Lemma: The permutation symbol obeys
$$ \varepsilon^{i_1 ... i_k} = \frac{1}{(n-k)!} g^{i_1 j_1} ... g^{i_k
j_k} \ \varepsilon_{j_1 ... j_n} \ \varepsilon^{j_{k+1} ... j_n} \,.\tag{11}$$
Proof of Prop.: Let $\eta := \mathrm{d}x^{i_1}\wedge...\wedge \mathrm{d}x^{i_k}$ and $\widetilde{\eta} := \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \mathrm{d}x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}$. In order that $\widetilde{\eta} = \star \eta$, we have to check if
$$ \forall \omega \in \Omega^k(M): \omega \wedge \widetilde{\eta} = ⟨\omega,\eta⟩ \ vol$$
Since both sides are linear in $\omega$, it suffices to check on a basis. Let $\omega := \mathrm{d}x^{1}\wedge...\wedge \mathrm{d}x^{k}$. Then, one immediately notices that $\eta = \varepsilon^{i_1 ... i_k} \omega$.
Observe that \begin{align*}
⟨\omega,\eta⟩ \ vol &= \varepsilon^{i_1 ... i_k} \ ⟨\omega,\omega⟩ \sqrt{g} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\
&= \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ g^{1 j_1} ... g^{k j_k} \ \varepsilon_{j_1 ... j_k} \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\
&= \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\
&= \sqrt{g} \ \Big( \frac{1}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \varepsilon^{j_{k+1} ... j_n} \Big) \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\
&= \mathrm{d} x^1 \wedge ... \wedge \mathrm{d} x^k \wedge \Big( \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \Big) \ \mathrm{d} x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}\\
&= \omega \wedge \widetilde{\eta}
\end{align*}
This concludes our proof. $\tag{Q.E.D.}$
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
$$
dz_1 \wedge d\bar{z}_3 \wedge d z_2 = - dz_1 \wedge dz_2 \wedge d\bar{z}_3.
$$
Best Answer
I think this page should be helpful: http://planetmath.org/hodgestaroperator