We transform the problem to a single variable question by the following

Lemma. Let $a_{1}, \ldots, a_{n}$ be nonnegative integers and let $G(x)$ be a nonzero polynomial with $\operatorname{deg} G \leqslant a_{1}+\ldots+a_{n}$. Suppose that some polynomial $F\left(x_{1}, \ldots, x_{n}\right)$ satisfies
$$F\left(x_{1}, \ldots, x_{n}\right)=G\left(x_{1}+\ldots+x_{n}\right) \quad \text{ for }\left(x_{1}, \ldots, x_{n}\right) \in\left\{0,1, \ldots, a_{1}\right\} \times \ldots \times\left\{0,1, \ldots, a_{n}\right\} \text{. }$$
Then $F$ cannot be the zero polynomial, and $\operatorname{deg} F \geqslant \operatorname{deg} G$.

For proving the lemma, we will use forward differences of polynomials. If $p(x)$ is a polynomial with a single variable, then define $(\Delta p)(x)=p(x+1)-p(x)$. It is well-known that if $p$ is a nonconstant polynomial then $\operatorname{deg} \Delta p=\operatorname{deg} p-1$.

If $p\left(x_{1}, \ldots, x_{n}\right)$ is a polynomial with $n$ variables and $1 \leqslant k \leqslant n$ then let
$$\left(\Delta_{k} p\right)\left(x_{1}, \ldots, x_{n}\right)=p\left(x_{1}, \ldots, x_{k-1}, x_{k}+1, x_{k+1}, \ldots, x_{n}\right)-p\left(x_{1}, \ldots, x_{n}\right)$$
It is also well-known that either $\Delta_{k} p$ is the zero polynomial or $\operatorname{deg}\left(\Delta_{k} p\right) \leqslant \operatorname{deg} p-1$.

Proof of the lemma. We apply induction on the degree of $G$. If $G$ is a constant polynomial then we have $F(0, \ldots, 0)=G(0) \neq 0$, so $F$ cannot be the zero polynomial.

Suppose that $\operatorname{deg} G \geqslant 1$ and the lemma holds true for lower degrees. Since $a_{1}+\ldots+a_{n} \geqslant \operatorname{deg} G>0$, at least one of $a_{1}, \ldots, a_{n}$ is positive; without loss of generality suppose $a_{1} \geqslant 1$.

Consider the polynomials $F_{1}=\Delta_{1} F$ and $G_{1}=\Delta G$. On the grid $\left\{0, \ldots, a_{1}-1\right\} \times\left\{0, \ldots, a_{2}\right\} \times \ldots \times\left\{0, \ldots, a_{n}\right\}$ we have
$$\begin{aligned} F_{1}\left(x_{1}, \ldots, x_{n}\right) & =F\left(x_{1}+1, x_{2}, \ldots, x_{n}\right)-F\left(x_{1}, x_{2}, \ldots, x_{n}\right)= \\ & =G\left(x_{1}+\ldots+x_{n}+1\right)-G\left(x_{1}+\ldots+x_{n}\right)=G_{1}\left(x_{1}+\ldots+x_{n}\right) \end{aligned}$$
Since $G$ is nonconstant, we have $\operatorname{deg} G_{1}=\operatorname{deg} G-1 \leqslant\left(a_{1}-1\right)+a_{2}+\ldots+a_{n}$. Therefore we can apply the induction hypothesis to $F_{1}$ and $G_{1}$ and conclude that $F_{1}$ is not the zero polynomial and $\operatorname{deg} F_{1} \geqslant \operatorname{deg} G_{1}$. Hence, $\operatorname{deg} F \geqslant \operatorname{deg} F_{1}+1 \geqslant \operatorname{deg} G_{1}+1=\operatorname{deg} G$. That finishes the proof.

To prove the problem statement, take the unique polynomial $g(x)$ so that $g(x)=\left\lfloor\frac{x}{m}\right\rfloor$ for $x \in\{0,1, \ldots, n(m-1)\}$ and $\operatorname{deg} g \leqslant n(m-1)$. Notice that precisely $n(m-1)+1$ values of $g$ are prescribed, so $g(x)$ indeed exists and is unique. Notice further that the constraints $g(0)=g(1)=0$ and $g(m)=1$ together enforce $\operatorname{deg} g \geqslant 2$.

By applying the lemma to $a_{1}=\ldots=a_{n}=m-1$ and the polynomials $f$ and $g$, we achieve $\operatorname{deg} f \geqslant \operatorname{deg} g$. Hence we just need a suitable lower bound on $\operatorname{deg} g$.

Consider the polynomial $h(x)=g(x+m)-g(x)-1$. The degree of $g(x+m)-g(x)$ is $\operatorname{deg} g-1 \geqslant 1$, so $\operatorname{deg} h=\operatorname{deg} g-1 \geqslant 1$, and therefore $h$ cannot be the zero polynomial. On the other hand, $h$ vanishes at the points $0,1, \ldots, n(m-1)-m$, so $h$ has at least $(n-1)(m-1)$ roots. Hence,
$$\operatorname{deg} f \geqslant \operatorname{deg} g=\operatorname{deg} h+1 \geqslant(n-1)(m-1)+1 \geqslant n$$