Let $F$ be the rank $n$ free group generated by $\phi_1,...,\phi_n$. Fix $m\geq 2$ integer and let $M$ be the subgroup of $F$ generated by $$\{\phi_i^m , [\phi_j,\phi_k] : 1\leq i,j,k\leq n \} $$ then $$F/M \cong \mathbb{Z}^n/m\mathbb{Z}^n$$
Is this true?
According to me, it's true beacuse the commutator subgroup is in $M$ then $F/M$ is abelian and by the relations $\phi_i^k$ we have the finiteness of the orders in each element, and this implies a "trivial" (the exponent of each $\phi_i$ in $F/M$ is send to the $i$-coordinate in $\mathbb{Z}^n/m\mathbb{Z}^n$) isomophism to $\mathbb{Z}^n/m\mathbb{Z}^n$.
Yes, it is indeed true, and I think you get the right idea of why it works, but it has to be written in a proper way.
Let us denote by $x_i$ the class of $\phi_i$ in $F/M$. Then $(x_1,\ldots,x_n)$ span the quotient. Using $x_ix_j=x_jx_i$ for all $i,j$ and $x_i^m=1$, you get that $F/M$ has at most $m^n$.
All is left to do is to construct a surjective morphism $f: F/M\to (\mathbb{Z}/m\mathbb{Z})^n$.
Indeed, assume that such $f$ exists. Then $F/M$ will have at least $m^n$ elements. By the previous step, $F/M$ has exactly $m^n$ elements. Since $f$ is surjective, it implies that $f$ is an isomorphism.
Now let $e_i=(0,0,\ldots,0,1,0,\ldots,0)$, where the $1$ is at the $i$-th position. Then $(e_1,\ldots,e_n)$ span $(\mathbb{Z}/m\mathbb{Z})^n$.
Since $F$ is free, there is a unique morphism $g:F\to (\mathbb{Z}/m\mathbb{Z})^n$ sending $\phi_i$ to $e_i$. Now $f([\phi_i,\phi_j])=[f(\phi_i,\phi_j])=0$ since $(\mathbb{Z}/m\mathbb{Z})^n$ is abelian, and $f(\phi_i^m)=mf(\phi_i)=0$ since $e_i$ has order $m$.
Hence $g$ factors through the set of relations to yield a morphism $f:F/M\to (\mathbb{Z}/m\mathbb{Z})^n$. This morphism is surjective because by definition ,$e_1,\ldots,e_n$ lie in its image.