$\newcommand\Legendre{\genfrac(){}{}}$We have $$\sum_{1\le i<j\le p}\Legendre{i-j}{p}x_{i}x_{j}\leq \sum_{1\le i<j\le p}x_{i}x_{j} = \frac{1}{2} \sum_{i \neq j} x_i x_j =\frac{1 - \sum_i x_i^2}{2} $$
and $$ \sum_{1\le i<j\le p}\Legendre{i-j}{p}x_{i}x_{j}= \frac{1}{2} \sum_{i \neq j} \Legendre{i-j}{p}x_{i}x_{j} \leq \frac{ \sqrt{p}}{2} \sum_i x_i^2$$
since the largest value of the quadratic form associated to a symmetric matrix on the unit sphere is its largest real eigenvalue, and the largest real eigenvalue of the $p \times p$ matrix with entries $\Legendre{i-j}{p}$ is the Gauss sum $1/p$.
A linear combination of $\sqrt{p}$ times the first inequality plus the second gives $$\sum_{1\le i<j\le p}\Legendre{i-j}{p}x_{i}x_{j}\leq \frac{1}{ 2 + 2 p^{-1/2}},$$ which is better than the claimed bound for $p> 16$.
It's not possible to improve this bound too much without getting a better bound than what's known for the clique number of the Paley graph, as we can take $x_i =1 /k$ for $i$ in a clique of size $k$ and $0$ otherwise