The statement in the question is false (as I mention in comments), but $(3)$ seems possibly to be what is meant.
For any positive $\{x_k\}$, Cauchy-Schwarz gives $ \left(\sum_{k=1}^nx_k\right)\left(\sum_{k=1}^n\frac1{x_k}\right)\ge n^2\tag{1} $ Let $\{r_k\}$ be the roots of $p$, then $ \left|\frac{a_1a_{n-1}}{a_0a_n}\right| =\left(\sum_{k_1}r_{k_1}\right)\left(\sum_{k_1}\frac1{r_{k_1}}\right) \ge\binom{n}{1}^2 $ $ \left|\frac{a_2a_{n-2}}{a_0a_n}\right| =\left(\sum_{k_1 $ \left|\frac{a_3a_{n-3}}{a_0a_n}\right| =\left(\sum_{k_1 $ \vdots\tag{2} $ Summing $(2)$ yields $ \sum_{i=1}^{n-1}\left|\frac{a_ia_{n-i}}{a_0a_n}\right|\ge\binom{2n}{n}-2\tag{3} $ If we include the end terms, we get the arguably more aesthetic $ \sum_{i=0}^n\left|\frac{a_ia_{n-i}}{a_0a_n}\right|\ge\binom{2n}{n}\tag{4} $
Note that $(3)$ and $(4)$ are sharp. If we cluster roots near $1$, we will get coefficients near $(x-1)^n$, for which the sums in $(3)$ and $(4)$ are equal to their bounds.