Can the $(n-1)$ power of the hermitian form on a compact complex $n$-fold be $\partial {\bar{\partial}}$-exact? Let $\omega$ be the hermitian form for an hermitian metric on a compact complex manifold.  Can $\omega^{n-1}$ be $\partial {\bar{\partial}}$-exact? (We know that our hermitian metric must necessarily be balanced and non-Kahler.) 
 A: Yes, this can happen.  For a simple example, consider $M = \mathrm{SL}(2,\mathbb{C})/\Lambda$ where $\Lambda\subset \mathrm{SL}(2,\mathbb{C})$ is a discrete, co-compact lattice.  Then $M$ is a compact complex $3$-manifold.
Let $\alpha_1,\alpha_2,\alpha_3$ be a basis for the right-invariant holomorphic $1$-forms on $\mathrm{SL}(2,\mathbb{C})$.  Because they are right invariant, they are well-defined on the quotient $M = \mathrm{SL}(2,\mathbb{C})/\Lambda$, and one can choose these forms so that
$$
\mathrm{d}\alpha_1 = \alpha_2\wedge\alpha_3\,,\quad
\mathrm{d}\alpha_2 = \alpha_3\wedge\alpha_1\,,\quad
\mathrm{d}\alpha_3 = \alpha_1\wedge\alpha_2\,.\quad
$$
The positive $(1,1)$-form
$$
\omega = \frac{i}{2}\left(\alpha_1\wedge\overline{\alpha_1}+\alpha_2\wedge\overline{\alpha_2}+\alpha_3\wedge\overline{\alpha_3}\right)
$$
defines an Hermitian structure on $M$ but is not closed (of course).  Meanwhile, we have
$$
\tfrac12\omega^2 = \tfrac14\left(
\alpha_2\wedge\alpha_3\wedge\overline{\alpha_2}\wedge\overline{\alpha_3}
+\alpha_3\wedge\alpha_1\wedge\overline{\alpha_3}\wedge\overline{\alpha_1}
+\alpha_1\wedge\alpha_2\wedge\overline{\alpha_1}\wedge\overline{\alpha_2}\right),
$$
while the formulae for the exterior derivatives of the $\alpha_i$ yield
$$
-i\,\partial\overline{\partial}\omega = \omega^2.
$$
