$\newcommand\Legendre{\genfrac(){}{}}$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 $\sqrt{p}$.
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 to get a value of $\frac{k-1}{ 2k}= \frac{1}{ 2 + 2 (k-1)^{-1}}$.
One can refine this slightly by replacing the second estimate with
$$ \sum_{1\le i<j\le p}\Legendre{i-j}{p}x_{i}x_{j}= \frac{1}{2} \sum_{i \neq j} \Legendre{i-j}{p}x_{i}x_{j} = \frac{1}{2} \sum_{i \neq j} \Legendre{i-j}{p}(x_{i}- 1/p)(x_{j}-1/p) \leq \frac{ \sqrt{p}}{2} \sum_i (x_i-1/p)^2 =\frac{ \sqrt{p}}{2} \left(\sum_i x_i^2 - 1/p\right) $$ which gives the upper bound
$$\sum_{1\le i<j\le p}\Legendre{i-j}{p}x_{i}x_{j}\leq \frac{p-1}{ 2p + 2 \sqrt{p}},$$ improving on the desired bound for $p> 9$.
