Let $(\frac{a}{p})$ denote the Legendre symbol. Let $P$ be a polynomial over $\mathbb{F}_p$ of degree $d$. I would like to upper bound
$$\Big\lvert\,\mathbb{E}_a \Big(\frac{P(a)}{p}\Big)\Big\rvert$$
where $a$ is chosen uniformly at random from $\mathbb{F}_p$.
Hasse's bound from elliptic curves seems to imply that when the degree of $P$ is 3, then
$$\Big\lvert\,\mathbb{E}_a \Big(\frac{P(a)}{p}\Big)\Big\rvert \le \frac{2}{\sqrt{p}}\;.$$
What can we say for higher degrees? Is it true that for any odd $d$,
$$\Big\lvert\,\mathbb{E}_a \Big(\frac{P(a)}{p}\Big)\Big\rvert \le \sqrt\frac{d+1}{p}\;?$$
As noted in my comment, when $P$ is a perfect square of sufficiently small degree (with respect to $p$), the conjecure is false.
Assuming $P$ is not a perfect square, we can find a bound, though not exactly what you conjectured. This answer is therefore incomplete.
Let $M$ be the number of distinct roots of $P$ in $\mathbb{F}_p$ and let $N$ be the number of $x\in\mathbb{F}_p$ such that $P(x)$ is a perfect square. The number of $a\in\mathbb{F}_p$ with $\left(\frac ap\right)=1$ is $N-M$ and the number of $a\in\mathbb{F}_p$ with $\left(\frac ap\right)=-1$ is $p-N$. Hence, $$ \mathbb{E}_a\left[\left(\frac{P(a)}a\right)\right] = \frac{2N-M-p}{p}. $$ Now, suppose $P$ is not a perfect square. Then $C:y^2=P(x)$ is an algebraic curve of degree $d$ and genus $g:=\frac12(d-1)(d-2)$. The number of affine points of $C$ is $2N$. Together with the point at infinity, we find $\#C(\mathbb{F}_p)=2N+1$. The generalized Hasse-Weil bound gives $$ \left|2N-p\right|<2g\sqrt{p}. $$ Now, the triangle inequality gives $$ \left|\mathbb{E}_a\left[\left(\frac{P(a)}a\right)\right]\right|\le \left|\frac{2N-p}p\right| + \frac{M}{p}<\frac{2g}{\sqrt{p}}+\frac{M}{p}=\frac{(d-1)(d-2)}{\sqrt p}+\frac{M}p. $$