Your sum is expressible as a linear combination of Jacobi sums. More precisely, let us notice that
$$
\left(\frac{x^3}{p}\right)=\left(\frac{x}{p}\right)
$$
for all $x$ and also that $(x+1)(x+\omega)(x+\omega^2)=x^3+1$. So, the sum $S$ in question is equal to
$$
S=\sum_{0\leq x\leq p-1}\left(\frac{x^3}{p}\right)\left(\frac{x^3+1}{p}\right)\chi_p(x^3).
$$
Now, let $\psi$ be one of two non-principal characters modulo $p$ with $\psi^3=\chi_0$. It is easy to see that
$$
\#\{x:x^3\equiv y \pmod p\}=1+\psi(y)+\psi^2(y)
$$
(i.e. it is $1$ when $y=0$, $3$ when $y$ is a nonzero cubic residue and $0$ if it is a cubic non-residue) Arranging the sum according to the value of $x^3$, we obtain
$$
S=\sum_{0\leq y\leq p-1}\left(\frac{y}{p}\right)\left(\frac{y+1}{p}\right)\chi_p(y)(1+\psi(y)+\psi^2(y)).
$$
To transform this into Jacobi sums, let us make a substitution $y=-z$ and notice that $\psi(-1)=1$, $\chi_p(-1)=-1$ and $\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}:=\varepsilon_p$. We get
$$
S=-\varepsilon_p\sum_z \left(\frac{z}{p}\right)\left(\frac{1-z}{p}\right)\chi_p(z)(1+\psi(z)+\psi^2(z)).
$$
Next, if $\eta_1,\eta_2$ are characters modulo $p$, then the Jacobi sum $J(\eta_1,\eta_2)$ is given by
$$
J(\eta_1,\eta_2)=\sum_z \eta_1(z)\eta_2(1-z).
$$
For instance, we get
$$
S=-\varepsilon_p(J(\chi_1,L_p)+J(\chi_2,L_p)+J(\chi_3,L_p)),
$$
where $L_p$ is the Legendre symbol and $\chi_i=\psi^iL_p \chi_p$. It is also known that if $\eta_1\eta_2$ is non-principal, then
$$
J(\eta_1,\eta_2)=\frac{g(\eta_1)g(\eta_2)}{g(\eta_1\eta_2)},
$$
(here $g$ is the Gauss sum $g(\eta)=\sum_n \eta(n)e^{2\pi i n/p}$)
so in this case we obtain
$$
S=-\varepsilon_p g(L_p)\sum_i \frac{g(\chi_i)}{g(\psi^i\chi_p)}
$$
and as a consequence of the identity $|g(\chi)|=\sqrt p$ for non-principal $\chi$, we prove Weil-like estimate
$$
|S|\leq 3\sqrt p.
$$
