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.) 



Can the $(n-1)$ power of the hermitian form on a compact complex $n$-fold be $\partial {\bar{\partial}}$-exact?
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
$$
-i\,\partial\overline{\partial}\omega = \omega^2.
$$


