Well, $N$ is generated by the commutators of elements of $\pi_1'(\overline{X})$ with elements of the maximal pro-$p$ subgroup of $\operatorname{Gal}(\mathbb F_q)$. This is just because any such element of $\operatorname{Gal}(\mathbb F_q)$ lies in the kernel of the map to the maximal prime-to-$p$ quotient of $\pi_1(X)$, and so its commutator with any element of $\pi_1(\overline{X})$ also lies in that kernel. Vice versa, modulo this subgroup, the action of the prime-to-$p$ quotient of $\operatorname{Gal}(\mathbb F_q)$ on $\pi_1(\overline{X})$ is well-defined, and so their semidirect product is clearly $\pi_1(X)'$. However this is not very explicit.

No, I think it is not really possible to give a reasonable presentation.

First let's show why the same problem is hard if we instead take $X$ to be an elliptic curve. Then the prime-to-$p$ fundamental group of $X$ is simply $\prod_{\ell\neq p} \mathbb Z_{\ell}^2$. On this, $\operatorname{Frob}_q$ acts with eigenvalues $\alpha,\beta$ that are the two roots of an polynomial with integer coefficients $x^2 - a x+ q$. $N$ contains the eigenspace of $\alpha$ if and only if the multiplicative order of $\alpha$ mod $\ell$ is not prime to $p$, and similarly for the eigenspace of $\beta$. This is because any more of $\operatorname{Frob}_q$ whose order is a power of $p$ is an element of the maximal pro-$p$ subgroup of $\operatorname{Gal}(\mathbb F_q)$, and its commutators are generated by all its eigenvectors with nontrivial eigenvalue.

This information does not really have a presentation in terms of generators and relations. It depends on the arithmetic of $\alpha,\beta$, and the number field they lie in.

Now let's show how this problem can be reduced to your problem.
Consider the case where $r=4$, $a_1=0,a_2=1,a_3=\infty, a_4=\lambda$ for some $\lambda \in \mathbb F_q$, $\lambda \neq 0,1$. Then the subgroup of $\pi_1'(\overline{X})$ generated by all products of two elements in the set $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ is the fundamental group of a degree $2$ cover of $x$ - it's the elliptic curve with equation $y^2=x(x-1)(x-\lambda)$ minus the points $x=0,x=1,x=\lambda,x=\infty$. Because the local monodromy generators at those four points are $\alpha_1^2,\alpha_2^2,\alpha_3^2,\alpha_4^2$, the quotient of that subgroup by the normal subgroup generated by $\alpha_1^2,\alpha_2^2,\alpha_3^2,\alpha_4^2$ is the fundamental group of this elliptic curve.

Given an explicit presentation of $N$, we can calculate a presentation for the intersection of $N$ with this index $2$ subgroup, and hence calculate the image of $N$ in this quotient. So the elliptic curve case can't be any harder than this case.

Using more tricks (Belyi's theorem) it is possible to reduce the problem for essentially any curve to the $r=3$ case. So I don't see this having a very easy solution.

However I think it is possible to calculate the image of $N$ in the maximal pro-$\ell$ quotient of $\pi_1(\overline{X})$. It is the whole pro-$\ell$ quotient if the multiplicative order of $q$ mod $\ell$ is divisible by $p$ and is trivial otherwise.

The reason is that $\operatorname{Frob}_q$ acts by multiplication by $q$ on the abelianization of the maximal pro-$\ell$ quotient (as you can see explicitly by looking at its action on the Galois groups of Kummer extensions). So in particular it acts as multiplication by $q$ mod $\ell$. Hence its commutator with elements of this subgroup is trivial mod $\ell$. so $N$ is a normal subgroup of a pro-$\ell$ group whose projection to the mod $\ell$ abelianization is the whole mod $\ell$ abelianization, and hence $N$ is the whole group.

On the other hand, if the order of $\operatorname{Frob}_q$ is prime to $p$, then every element of $\operatorname{Gal}(\mathbb F_q)$ of order a power of $p$ acts trivially on the mod-$\ell$ abelianization of this pro-$\ell$ group. Hence, by the Hall-Burnside theorem, it acts trivially on the whole pro-$\ell$ group. So there are no nontrivial commutators.