Examples of compact complex non-Kähler manifolds which satisfy $h^{p,q} = h^{q,p}$ The existence of a Kähler metric on a compact complex manifold $X$ imposes restrictions on it's Dolbeault cohomology; namely, $h^{p,q}(X) = h^{q,p}(X)$ for every $p$ and $q$. I am looking for some explicit examples of compact complex manifolds which satisfy these restrictions, but are not Kähler (i.e. do not admit a Kähler metric).
In dimension 1, every compact complex manifold is Kähler ($d\omega$ is automatically zero). 
In dimension 2, a compact complex surface $X$ is Kähler if and only if the first Betti number $b_1(X)$ is even. Therefore, if $h^{1,0}(X) = h^{0,1}(X)$, then $b_1(X) = h^{1,0}(X) + h^{0,1}(X) = 2h^{1,0}(X)$ so by the aforementioned result, $X$ is Kähler.
It follows that any examples must be of dimension at least three. 
The only thing that has come to mind so far is Hironaka's example which is a deformation of Kähler manifolds such that the limiting fibre is not Kähler. As diffeomorphism type is preserved under deformations, I know that the limiting fibre has the same Betti numbers as a Kähler manifold (i.e. the odd Betti numbers are even), but I don't know if its Hodge numbers satisfy the desired symmetry.
Added Later: 


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*After consulting the book Cohomological Aspects in Complex Non-Kähler Geometry by Angella, I realised that the central fibre of Hironaka's deformation, which I will now call $X_0$, is indeed an example. We just need to show that $h^{p,q}(X_0) = h^{q,p}(X_0)$.
First note that $X_0$ is a Moishezon manifold and is therefore a proper modification of a projective manifold. As projective manifolds satisfy the $\partial\bar{\partial}$-lemma, so do Moishezon manifolds; see Theorem 5.22 of Real Homotopy Theory of Kähler Manifolds by Deligne, Griffiths, Morgan, and Sullivan. In the same paper, they show that if a compact complex manifold satisfies the $\partial\bar{\partial}$-lemma, the Hodge-Frölicher spectral sequence degenerates at the first step, see Theorem 5.21. This in turn equivalent to the equality (for each $k$) $$b_k(X_0) = \displaystyle\sum_{p+q = k}h^{p,q}(X_0),$$ so $$b_k(X_a) = \displaystyle\sum_{p+q = k}h^{p,q}(X_a)$$ for all $a$ in the base of the Hironaka deformation. The map $a \mapsto h^{p,q}(X_a)$ is upper semicontinuous in general (see Corollary 9.19 of Voisin's Hodge Theory and Complex Algebraic Geometry I for example), but the map $a \mapsto b_k(X_a)$ is constant by Ehresmann's result. Due to the above equality, we see that $a \mapsto h^{p,q}(X_a)$ is actually constant for all $p$ and $q$. Therefore, $$h^{p,q}(X_0) = h^{p,q}(X_a) = h^{q,p}(X_a) = h^{q,p}(X_0)$$ where $a \neq 0$ and we use the fact that $X_a$ is projective and hence Kähler.
The above result can be summarised as follows:

Let $\mathcal{X} \to B$ be a deformation of compact complex manifolds with $0 \in B$. If $a \mapsto h^{p,q}(X_a)$ is constant on $B\setminus\{0\}$, and the Hodge-Frölicher spectral sequence of $X_0$ degenerates at the first step, then $a \mapsto h^{p,q}(X_a)$ is constant on $B$.

In particular, given a deformation $\mathcal{X} \to B$ such that $h^{p,q}(X_a) = h^{q,p}(X_a)$ for all $a \in B\setminus\{0\}$, and the Hodge-Frölicher spectral sequence of $X_0$ degenerates at the first step, then $h^{p,q}(X_0) = h^{q,p}(X_0)$. If $X_0$ is not Kähler (as in Hironaka's example), $X_0$ is then an example of the type I was searching for.

*If $h^{p,q} = h^{q,p}$ for all $p$ and $q$, then $b_k$ is even for $k$ odd. So one may instead try to solve the easier problem of finding a non-Kähler manifold $X$ such that $b_k(X)$ is even for $k$ odd. 
As above, in the one or two dimensional case, if $b_k$ is even for $k$ odd, then $X$ is Kähler. In dimension three however, there is a simple example given by $S^3\times S^3$. A reference for the fact that $S^3\times S^3$ can be equipped with a complex structure is Complex Structures on $S^3\times S^3$ by Fujiki. 
The Kunneth formula gives $b_0 = 1$, $b_1 = 0$, $b_2 = 0$, $b_3 = 2$, $b_4 = 0$, $b_5 = 0$, and $b_6 = 1$. All odd Betti numbers are even, but $S^3\times S^3$ is not Kähler as $b_2 = 0$. In fact, all even Betti numbers of a Kähler manifold are positive ($\omega^k$ is not exact), so one might try to find an example of a non-Kähler manifold with $b_{2k} > 0$ and $b_{2k+1}$ even. By the discussion in point one above, an example of such a manifold is the central fibre of the Hironaka deformation.
 A: I believe (he said with some trepidation) that the results of Angella and Tomassini (which say that this property is preserved under complex analytic deformations) give lots of examples. This is in Angella's thesis, and in their joint Inventiones article (2012) - they also discuss at length the obstructions (or lack thereof) to the existence of such deformations.
A: Complete smooth toric varieties $X_\Sigma$, corresponding to regular non-polytopal fans $\Sigma$ form a large series of manifolds bimeromorphic to Kahler, but non-Kahler.
At the same time, their Hodge diamond is concetrated on the main diagonal: $h^{p,q}=0$, unless $p=q$.
The smallest possible dimension of such an example is 3: see Fulton's "Introduction into toric varieties", Excersise in Section 3.4.
A: Every compact complex manifold satisfying the $\partial\overline{\partial}$-Lemma has such a property. Particular examples are given by - as you already said - Hironaka example (and, more in general, Moishezon manifolds and manifolds in class C of Fujiki), or some deformations of twistor spaces (see LeBrun, Poon, Twistors, Kähler manifolds, and bimeromorphic geometry. II, J. Amer. Math. Soc. 5 (1992), no. 2, 317–325).
On the other side: in Ceballos, Otal, Ugarte, Villacampa, arXiv:1111.5873, Proposition 4.3, you find a concrete example of a compact complex manifold with the symmetry of Hodge diamond you require. This example does not satisfy $\partial\overline{\partial}$-Lemma.
A: Take $\mathbb{C}^2\setminus \{0\}/\mathbb{Z}$ where the equivalent relationship takes $(z_1,z_2)\sim \lambda(z_1,z_2)$ with $\lambda\neq0$. This manifold is $S^1\times S_3$ and it's definitely complex (comes from $\mathbb{C}^2$ anyway). It's compact (very obvious), but it's not Kähler, because $H^{2}(S^1\times S^3)=H^0(S^1)\otimes H^2(S^3)\oplus H^1(S^1)\otimes H^1(S^3)\oplus H^2(S^1)\otimes H^0(S^3)=0$. So there doesn't exist any closed (1,1)-form. 
