Oops, I was thinking of a Lie group. Edit: On a Lie group, it is pretty easy to test for integrability. Take a basis of complex linear left invariant 1-forms (i.e. left translate from a choice of such 1-forms on the Lie algebra). Then compute their exterior derivatives. You have an integrable almost complex structure if and only if the (0,2) parts of the exterior derivatives all vanish. To see this, you express the exterior derivatives in linear combinations of the wedge products of the original 1-forms and their complex conjugates. So examples are easy to check, and only require the differential familiar from Lie algebra cohomology, i.e. a pure Lie algebra calculation.
On a homogeneous space $G/H$ the computation is a little trickier. You take $\omega=g^{-1}dg$, the left invariant Maurer Cartan form on $G$, and then $\omega+\mathfrak{h}$ is semibasic for the quotient map $G \to G/H$ and splits into a complex linear part and a conjugate linear part . on each tangent space of $G/H$. Let $\eta$ be the complex linear part. Pick a splitting of $\mathfrak{g}$ into $\mathfrak{g}/\mathfrak{h}$ and some complement, identified with $\mathfrak{h}$, and let $\Omega$ be the projection of $\omega$ to complement. The equation $d \omega + (1/2)[\omega,\omega]=0$ gives an equation $d \eta = - \rho(\omega) Omega \wedge \eta + a \eta \wedge \eta + b \eta \wedge \bar{\eta} + c \bar{\eta} \wedge \bar{\eta}$. The Nijenhuis tensor is vanishes just when $c$. c=0$. I am pretty sure that the generic invariant almost complex structure on a flag manifold (invariant under the compact form of the automorphism group) is not integrable, but I would have to check.
