I just consider the case $h=3$, but the argument is completely general.
From the classification of closed surfaces, we know that $N_3$ is the connected sum of three projective spaces, so that $N_3$ be the quotient of an hexagon $P$ by identifying its sides as indicated by the following figure:
Now let $U, V \subset N_3$ be subspaces illustrated by the figure below:
Applying van Kampen theorem, we deduce that $\pi_1(N_3)$ is the quotient of the free product $\pi_1(U) \ast \pi_1(V)$ identifying the images of $\pi_1(U \cap V)$ into $\pi_1(U)$ and $\pi_1(V)$. In fact, $\pi_1(V)$ is clearly trivial so that $\pi_1(N_3)$ turns out to be the quotient of $\pi_1(U)$ by its subgroup corresponding to $\pi_1(U \cap V)$.
Now, there is naturally a deformation retract $\phi$ from $U$ to the quotient of $\partial P$: it is just a graph $\Gamma$, more precisely a bouquet of three circles labelled by $a_1$, $a_2$ and $a_3$. Therefore, $\pi_1(U)$ is just the free group $\langle a_1,a_2,a_3 \mid \ \rangle$.
Then, there is a deformation retract from $U \cap V$ to the circle $\partial V$. Therefore, $\pi_1(U \cap V)$ is infinite cyclic and its image into $\pi_1(U)$ is the homotopy class of the circle $\phi(\partial V)$ in the quotient $\Gamma$ of $\partial P$. Such a loop in $\Gamma$ is labelled by $a_1^2a_2^2a_3^2$.
Finally, we conclude that $$\pi_1(N_3)= \langle a_1,a_2,a_3 \mid a_1^2a_2^2a_3^2 \rangle.$$