Consider a degree $n$ polynomial $P(x)$ with coefficients $c_i \in \{-1,0,1\}$ chosen uniformly and independently.
What is the probability that $P(x)$ has a root which is a root of unity?
For a particular polynomial you can test the property by just seeing if it is divisible by $x^k-1$ for some $k$.
This is not meant to be an answer, it's more of a "too big comment", but you can use the following as a lower bound : we have $p(1) = 0$ if and only if $\sum_{i=0}^{n-1} c_i = -1$ (assuming $p(x) = x^n + \sum_{i=0}^{n-1} c_i$), and using a counting argument, you get $$ \mathbb P \ge \frac{ \sum_{d=0}^{\lfloor \frac{n-1}2 \rfloor} \binom{n}{d,d+1,n-2d-1}}{3^n}. $$
(you want $d$ of the $c_i$'s to have value $1$, $d+1$ of them to have the value $-1$ and the other ones $0$, $d$ ranging from $0$ to $\lfloor (n-1)/2 \rfloor$). You can compute this lower bound probably more explicitly if you use techniques from random walks, i.e. you want $\sum_{i=0}^{n-1} c_i = -1$, this is saying that the Markov chain starting at $0$ and doing steps of length $1$ in $\mathbb Z$ ends up at $-1$ after precisely $n$ steps.
This is only a lower bound because I only considered the root of unity $1$. You could get a bigger lower bound using similar techniques with $-1$ and maybe if you push a bit more with $i$ and $\frac{\pm 1 \pm \sqrt{-3}}2$, since these roots of unity generate lattices in $\mathbb C$ over which you can perform random walks. But this is going nowhere close to an answer, so it should be left aside.
I must admit the problem looks tough, I'm looking forward to a great answer.