I don't think it follows the $$4n + 2$$ rule. The double bonds give 8 π electrons. Then the lone pairs on the $$\ce{NH}$$ and $$\ce{NH2}$$ groups are delocalized (are they?) so they must contribute 4 π electrons. Thus in total there are 12 π electrons, which doesn't follow Hückel's rule. So, why is it aromatic?
• Hückel's rule is for conjugated circuits, so $\ce{NH2}$ group is irrelevant, whereas $\ce{NH}$ is (lone pair on $\mathrm{2p}_z$ orbital), resulting in $n = 10$ and overall aromaticity. – andselisk May 10 '19 at 5:45
The $$4n+2$$ rule applies only for a conjugated ring. If you assume that the amino group contributes a pair of pendant $$\pi$$ electrons into the ring system then perforce you do not have a conjugated ring. And even if you don't assume such a delocalization from the amino group, the cross-link within the ring system is still a stretch for the $$4n+2$$ rule.