I did not check all your arguments but indeed $\nabla_X(\star_0 1) = 0$ in your setting. To see it, choose $p \in M$ and a local orthonormal frame $(E_1,\dots,E_d)$ on a neighborhood $U$ of $p$. Then,
$$ (\nabla_X(\star_0 1))(p) = \nabla_X(E_1 \wedge \dots \wedge E_d) = \sum_{i=1}^d E_1 \wedge \dots \wedge E_{i-1} \wedge \nabla_X E_i \wedge E_{i+1} \wedge \dots \wedge E_d. $$
Write $\nabla = D + A$ on $U$ where $D$ is the trivial connection and $A = (\omega^i_j) $ is a $d \times d$ matrix of one-forms (the connection form) defined by $\nabla_X E_i = \omega_i^j(X) E_j$. Since $\nabla$ is metric, the matrix $D$ is anti-symmetric and thus we have
$$ \sum_{i=1}^d E_1 \wedge \dots \wedge E_{i-1} \wedge \nabla_X E_i \wedge E_{i+1} \wedge \dots \wedge E_d = \\
\sum_{i,j=1}^d E_1 \wedge \dots \wedge E_{i-1} \wedge \omega^j_i(X)E_j \wedge E_{i+1} \wedge \dots \wedge E_d = \\
\sum_{i=1}^d E_1 \wedge \dots \wedge E_{i-1} \wedge \omega^i_i(X)E_i \wedge E_{i+1} \wedge \dots \wedge E_d = \left( \sum_{i=1}^n \omega^i_i(X) \right) E_1 \wedge \dots \wedge E_d = 0$$
as $\omega^i_i \equiv 0$.
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
Yes, the relation you wrote is true. The Hodge star on any manifold $M$ is a pointwise operator and is $\mathbb{R}$-linear at each point $p$ (i.e., $\ast_p: \Lambda^{\cdot} T_p^\ast M \to \Lambda^{\cdot} T_p^\ast M$ is a linear map between vector spaces). This means that for any differential form $\omega$ and any smooth function $h$, at each point $p$, $$(\ast(h \omega))(p) = \ast_p (h(p) \omega(p)) = h(p) \ast_p(\omega(p)) = h(p)\ast(\omega) (p).$$ (I hope my notation is clear.) Thus $\ast(h\omega) = h \ast(\omega)$.
Said in fancy terms, this property of $\ast$ means that it's a vector bundle homomorphism, or a $C^\infty(M)$-module homomorphism.