I'm not sure if I understand just what you're asking, but perhaps it is the following:

Suppose we know $a_0$ and $a_{n-1}$, and we know that $f(x)$ has only integer roots. Can we use this information to meaningfully bound the coefficients?

If that's what you're asking, then the answer is no. Consider for example the polynomial
$$
g(x) = x (x^2 - c^2)^{k},
$$
where $c \in \mathbb{Z}$. Then this has $a_0 = a_{n-1} = 0$. But even for fixed $n$, the other coefficients are unbounded if you let $c$ be large.

## This part added in response to Timothy Chow's comment:

Timothy Chow made the suggestion that allowing $a_{0} = a_{n-1} = 0$ feels like cheating, and on reflection, I agree!

**If we allow $a_0 = 0$, then we can't get any bound on the other coefficients**

Namely, just consider any monic degree $n-2$ polynomial of the form $p(x) = x^{n-2} + b_{n-4} x^{n-4} + b_{n-5} x^{n-5} + \cdots + b_0$. Then look at the polynomial $x (x+a) p(x) = x^n + a x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + 0$. This will have integer roots, and the coefficient $a_{n-1}$ is free to be any integer you like. Yet the other coefficients can be wild. For concreteness, consider this
$$
h(x) = x (x+a)(x^2 - c^2)^k,
$$
which, for fixed $a$ and $k$, will have unbounded coefficients if you pick $c$ large enough [e.g., the sum of these coefficients is $h(1)$ which is unbounded].

**If we disallow $a_0 = 0$, then we do get a bound**

However! If we disallow the case $a_0 = 0$, then *yes* you can get a bound on the remaining coefficients. Namely, if $f(x)$ factors as $f(x) = \prod_{i} (x +r_i)$, then we know $a_0 = \prod_{i} r_i$. If none of the roots is $0$ (i.e., if $a_0 \neq 0$), then each root is at least $1$ in absolute value, and thus $|a_0|$ is an upper bound on every product of the form $\prod_{i \in S} |r_i|$. Therefore, we have

$$
|a_{n-k}| = \left| \sum_{|S| = k} \prod_{i \in S} r_i \right| \leq \sum_{|S| = k} \prod_{i \in S} |r_i| \leq \sum_{|S| = k} |a_0| = {n \choose k} |a_0|.
$$
This bound is tight of course by the polynomials $(x+1)^n$ and $(x-1)^n$. (This reminds me of Newton's inequalities, which probably can provide an alternate proof.)