Largest absolute value of a polynomial of degree $n$ on $\{0,1,\ldots,n\}$

You can write your polynomial as $$ P(x) = \sum_{k = 0}^n P(k) L_{n,k}(x) ,$$ where $L_{n,k}$ are the Lagrange interpolation polynomials with nodes at $0, 1, \ldots, n$. Note that $(-1)^{n - k} L_{n,k}$ has positive coefficient at $x^n$. Thus, with the constraint $$\max\{|P(k)| : k = 0, 1, \ldots, n\} \leqslant 1,$$ the largest possible value of the coefficient of $P$ at $x^n$ is attained by $$ \bar P(x) = \sum_{k = 0}^n (-1)^{n - k} L_{n,k}(x) , $$ and the coefficient of $\bar P$ at $x^n$ is equal to $$ \sum_{k = 0}^n \prod_{\substack{0 \leqslant j \leqslant n \\ j \ne k}} \frac{1}{|k - j|} \, .$$ In other words, if the coefficient at $x^n$ is to be equal to $1$, the least possible value of $\max\{|P(k)| : k = 0, 1, \ldots, n\}$ is $$ \biggl(\sum_{k = 0}^n \prod_{\substack{0 \leqslant j \leqslant n \\ j \ne k}} \frac{1}{|k - j|} \biggr)^{-1} .$$ It remains to note that $$ \sum_{k = 0}^n \prod_{\substack{0 \leqslant j \leqslant n \\ j \ne k}} \frac{1}{|k - j|} = \sum_{k = 0}^n \frac{1}{k! (n - k)!} = \frac{1}{n!} \sum_{k = 0}^n \binom{n}{k} = \frac{2^n}{n!} \, .$$

Lagrange interpolation suggested by Mateusz Kwaśnicki is perfectly ok, but in this case it is probably easier to use the finite difference formula $$\sum_{i=0}^{n} (-1)^{i}{n\choose i}P(t+n-i)=\Delta^n P=n!$$ for $t=0$, where $\Delta:f(t)\to f(t+1)-f(t)$ is a finite difference operator.

Also this very statement is well known, in case if you need references.