The nitrogen with a methyl group does indeed contribute to aromaticity in this molecule. The criteria for aromaticity under Hückel's rule are that $4n+2$ electrons reside in coplanar p-orbitals in a cyclic system. This allows the p-orbitals to overlap and form a set of π-orbitals, which are delocalised over the whole ring system.
In this example all the atoms in the ring are sp2 hybridised. The three carbons in the ring form three σ bonds with their sp2 orbitals and then have one half-filled p-orbital perpendicular to the plane of the ring which participates in π-bonding.
The nitrogen with a double bond forms two σ bonds with two of its sp2 orbitals and uses the third to accommodate its lone pair. The lone pair is therefore in the plane of the ring, facing outwards, and is thus unable to participate in π-bonding. It is left with a half-filled p-orbital in the plane which contributes another electron to the π-system.
The nitrogen with the methyl group attached forms three σ bonds with its sp2 orbitals, and therefore has a full p-orbital remaining perpendicular to the plane of the ring, which can overlap with the other p-orbitals to form a π-system.
In total there are six π-electrons residing in five coplanar p-orbitals. So, Hückel's rule is fulfilled and aromaticity is present.