Let $u_k$ be the number of variables with value exactly $k$. If you pick a distribution of the $u_k$ such that $E[u_k]=1$, $E[u_k^2] = 2-1/n$, $E[u_ku_l] = 1-1/n$ for $k \neq l$ and $\sum_{k=1}^n u_k$ is always $1$, then by choosing a random $u_1, \dots ,u_n$ according to the distribution and then choosing a random ordering, you get a random set of variables with the desired independence property.
On the other hand if you do the same thing for $n+1$ pairwise independent random variables you get $E[u_k]=1+1/n$, $E[u_k^2] = 2+ 2/n$, $E[u_k u_l] = 1+1/n$. Here is a way to construct a distribution on $u_k$ that satisfies these conditions. To get a distribution for $n$ pairwise independent random variables, we just delete the last variable. The formula is:
With probability $(1+1/n) /k(k+1)$, we have $u_k=k+1$, $u_l =0$ for $l <k$, $u_l=1$ for $l>k$. To compute the three moments, we ignore a factor of $1+1/n$:
Mean:
$$ \frac{(k+1)}{k(k+1)} + \sum_{l < k} \frac{1}{l(l+1) } = \frac{1}{k} + 1- \frac{1}{k} = 1 $$
Mean of square:
$$ \frac{(k+1)^2}{k(k+1)} + \sum_{l < k} \frac{1}{l(l+1) } = \frac{k+1}{k} + 1- \frac{1}{k+1} = 2 $$
Mean of product is the same as mean, since if $l<k$ then $u_l=1$ whenever $u_k>0$.
Then put back in the factor of $(1+1/n)$ to get the right answer.
The expected value of the minimum of the $n+1$ variables at least $\sum_{k=1}^n (1+1/n) /(k+1) \approx \log n$. Removing a variable won't make the minimum any smaller.
Best Answer
As discussed in comments, I think for large $n$ the probability that it has a root which is a root of unity is double the probability that 1 is a root. For large $n$, $P(1)$ is a random variable whose distribution is approximately normal and whose variance is $\sigma^2=2n/3$. The probability that $P(1)=0$ is then approximately $1/\sigma\sqrt{2\pi}$. Doubling this gives a probability $\sqrt{3/\pi n}\approx 0.98 n^{-1/2}$. This seems to agree quite well with the final two values in Matt F.'s list:
$$\frac{263}{729}=0.361 \qquad \sqrt{\frac{3}{7\pi}}=0.369$$
$$\frac{2267}{6561}=0.346 \qquad \sqrt{\frac{3}{8\pi}}=0.345$$