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If enough Hodge numbers vanish so that the Hodge structure $H^{2k+1}(X)$ has level one, then $J^kX$ should be an abelian variety. This applies to Fano (e.g. cubic) 3-folds for example.
Later that day: Partly in response to Charles Siegel's comment/question, let me sketch a proof of a slightly more general statement. Suppose X is a projective rather than just Kaehler (which I forgot to say before), so $H$ has a polarization $Q$. Assume further that $$H^{2k+1}(X) = H= H^{pq}\oplus H^{qp}$$ has only two terms. Let $G$ be the same thing as $H$ viewed as a weight one structure. More precisely, the lattices $G_Z=H_Z$ are the same, and $G^{10}=H^{pq}$.
Then one sees that $J^kH= G^{01}/G_Z$, and that $\pm Q$ gives a polarization on $G$. So this is abelian variety.
If enough Hodge numbers vanish so that the Hodge structure $H^{2k+1}(X)$ has level one, then $J^kX$ should be an abelian variety. This applies to Fano (e.g. cubic) 3-folds for example.