This is a standard question in diophantine approximation. See for example Chapter 3 of Waldschmidt's book Diophantine Approximation of Linear Algebraic Groups. Here is the simplest bound that Waldschmidt gives. Assuming $a_0\ne 0$, let $H := \max_{0\le i\le n} |a_i|$. We will show that if $\alpha$ is any nonzero root of your polynomial, then $|\alpha| > 1/(H+1)$.

If $|\alpha| \ge 1$ then the inequality trivially holds. Otherwise, if $|\alpha|<1$, then

$${1\over |\alpha|} \le \left|{a_0\over \alpha}\right| = \left| a_1 + a_2\alpha + \cdots + a_n \alpha^{n-1} \right| \le H(1+|\alpha| + \cdots+|\alpha|^{n-1}) < {H\over 1-|\alpha|},$$ which rearranges to $|\alpha| > 1/(H+1)$.

This is a standard question in diophantine approximation. See for example Chapter 3 of Waldschmidt's book Diophantine Approximation of Linear Algebraic Groups. Here is the simplest bound that Waldschmidt gives. Assuming $a_0\ne 0$, let $H := \max_{0\le i\le n} |a_i|$. We will show that if $\alpha$ is any nonzero root of your polynomial, then $|\alpha| > 1/(H+1)$.

If $|\alpha| \ge 1$ then the inequality trivially holds. Otherwise, if $|\alpha|<1$, then

$${1\over |\alpha|} \le \left|{a_0\over \alpha}\right| = \left| a_1 + a_2\alpha + \cdots + a_n \alpha^{n-1} \right| \le H(1+|\alpha| + \cdots+|\alpha|^{n-1}) < {H\over 1-|\alpha|},$$ which rearranges to $|\alpha| > 1/(H+1)$.

Edit: See David Lampert's comment below for how to use the above bound to answer the question that David Harris asked.