Even though this question has an accepted answer, the answers so far are not complete or explicit. I kept working on this question, because I have been curious for a long time about the structure of Dolbeault complexes. First of all, the Frölicher spectral sequence does not directly reveal all of the non-Hodge information in the Dolbeault complex of a non-Kähler complex manifold. I learned from Mikhail Khovanov that in the category of bounded double complexes over a field, every object is isomorphic to a unique direct sum of indecomposable objects. Moreover, the indecomposable double complexes can be classified as squares, dots, and zigzags. Here is an example of each type of indecomposable, with the convention that omitted cells and arrows are 0:
$$\begin{matrix} \mathbb{C} & \rightarrow & \mathbb{C} \\\\
\uparrow & & \uparrow \\\\
\mathbb{C} & \rightarrow & \mathbb{C} \end{matrix}\qquad\qquad
\mathbb{C}\qquad\qquad
\begin{matrix} \mathbb{C} & \rightarrow & \mathbb{C} \\\\
& & \uparrow \\\\
& & \mathbb{C} & \rightarrow & \mathbb{C} \end{matrix} $$
This is actually a standard result about an $A_\infty$ quiver algebra with alternating arrows. To be precise about the Hodge theorem and the Frölicher spectral sequence, the $\partial\bar{\partial}$ lemma says exactly that there are no zigzags (other than length 1, which are then dots), which is then equivalent to the statement that horizontal cohomology is isomorphic to vertical cohomology. The squares are projective objects and do not contribute to any cohomology theory. The odd-length zigzags contribute to the total de Rham cohomology, while the even-length zigzags do not. Meanwhile there are two Frölicher spectral sequences, each of which detects half of the even zigzags. The Frölicher spectral sequences are insensitive to the odd zigzags, but you can still say that a Dolbeault complex is non-Hodge if it has odd zigzags, even though each Frölicher spectral sequence degenerates at $E_1$ if there are no even zigzags.

The information of all of the zigzags, extracted by discarding only the squares, has been defined in the literature as "Aeppli cohomology". (Serre duality implies, sort-of indirectly, that the Aeppli cohomology of a compact manifold is self-dual; I would be interested to see a direct derivation of this fact.)

Now, the Hopf manifolds. The most standard round Hopf manifold of complex dimension $n$ has an important group action of $U(n) \times S^1$. (For those who aren't familiar with the terminology, the standard Hopf manifold is $(\mathbb{C}^n\setminus 0)/\Gamma_r$, where $\Gamma_r$ is generated by rescaling by a real constant $r > 1$.) There is one aspect of my calculation that for me is a conjecture: That a connected, compact Lie group acts trivially on all of the zigzags of a compact manifold. This is true for odd zigzags since the de Rham cohomology has an invariant integer lattice, but I do not have an argument for even zigzags. But let's suppose that it is so.

The invariant part of the Dolbeault complex is algebraically generated by these differential forms of degree $(1,0)$, $(0,1)$, and $(1,1)$:
$$\alpha = \frac{\bar{z} \cdot dz}{z \cdot \bar{z}} \qquad
\bar{\alpha} = \frac{z \cdot d\bar{z}}{z \cdot \bar{z}}
\qquad \omega = \frac{dz \cdot d\bar{z}}{z \cdot \bar{z}}.$$
(I use $z$ and $dz$ as a vector of functions and a vector of 1-forms, so that I can take dot products. I'm leaving out the wedge product symbol.) Then I calculated the following:
$$\partial \alpha = 0 \qquad \bar{\partial} \alpha = \alpha \bar{\alpha} - \omega \qquad \partial \omega = - \alpha \omega \qquad \omega^n = n\alpha \bar{\alpha} \omega^{n-1}.$$
A basis for the invariant part of the Dolbeault complex is given by $\omega^k$, $\alpha \omega^k$, $\bar{\alpha} \omega^k$, and $\alpha \bar{\alpha} \omega^k$ for $0 \le k \le n-1$. The Poincaré series of the invariant complex is a matrix like this one:
$$\begin{matrix} 0 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 1 & 2 & 1 \\\\ 0 & 1 & 2 & 1 & 0 \\\\ 1 & 2 & 1 & 0 & 0 \\\\ 1 & 1 & 0 & 0 & 0 \end{matrix}.$$
After calculating the differential, my answer is that this decomposes as a dot at each corner, a zigzag of length 3 next to each each corner, and a progression of squares. Modulo the conjecture that all zigzags are invariant, this is a complete description of the Dolbeault complex. The $n=1$ case is an exception in which the Hopf manifold obviously is Kähler (it's a torus, and all complex curves are Kähler). In this case the invariant complex decomposes as four dots.