Non-Kahler manifolds and the dd^c-lemma Is there an explicit example of non-Kahler manifold $M$ such that $M$ satisfies the dd^c lemma ?
 A: This does not answer your question, but it seems interesting to point out.
A non-example would be any complex threefold diffeomorphic to the six-sphere (if such a manifold exists).
To see this, first note that the $dd^c$-lemma holds for $(1,1)$-forms on a compact complex manifold iff $b_1=2h^{0,1}$ (see for example the paper of Paul Gauduchon "La 1-forme de torsion d'une variete hermitienne compacte", Math.Ann. 1984, page 504).
Now any threefold diffeomorphic to the six-sphere must have $h^{0,1}\geq 1$ by a result of Alfred Gray ("A property of a hypothetical complex structure on the six-sphere", Bollettino U.M.I. 1997, Theorem 5), and therefore the $dd^c$-lemma for $(1,1)$-forms fails since $b_1=0$.
A: Here is an example of a Moishezon manifold which is easy to visualize. Take a high degree (e.g. a quintic) hypersurface $Z$ in $\mathbb{P}^{4}$ which has a single ordinary double point. Let $X$ be a small resolution of $Z$. Explicitly, a small analytic neighborhood of the singularity can be identified with the vertex of a cone over a two dimensional quadric and you just need to blow-up the Weil divisor which is the preimage of one ruling. The threefold $X$ is compact complex manifold and does not admit any Kaehler structure. The last statement follows for instance from a theorem of Smith-Thomas-Yau which states that a threefold with a single node will admit a symplectic small resolution only if the three sphere that vanishes at the node is homologous to zero. The high degree condition on $Z$ ensures that the vanishing cycle is not homologous to zero, hence the statement.  
A: I think this is an example.
There is an integrable complex structure on X=S^3 xS^3.  For topological reasons we have
Td(X)[X]=1.  Thus h^0,1 =0.  It's an easy exercise that a complex manifold with h^0,1 =0 satisfies the d\bar{d} lemma.
