The answer is $\boxed{c = \frac{4}{3}}$. To see this, take $n$ divisible by $9$, where $\frac{8}{9}$ of the numbers are $-1$ and the remainings are $2$. Otherwise, the inequality is strict. Now, notice that \[ \sum_{k = 1}^n (x_k + 1)(x_k - 2)^2 \ge 0 \Rightarrow \sum_{k = 1}^n 3x_k^2 \le \sum_{k = 1}^n x_k^3 + 4 = 4n \]This proves $\sum_{k = 1}^n x_k^2 \le \frac{4}{3} n$, which is what we wanted.

Here is braindead calculus bash Define functions $g:\mathbb{R}^n\rightarrow \mathbb{R}^n, f: A \rightarrow \mathbb{R}^n$ where $g(x_i)= \sum_{i=1}^n x^3_i ,A=\{\mathbb{R} \ni x_1,x_2,...,x_n \ge -1| g(x_i)=0 \}$, $f(x_i)= \sum_{i=1}^n x_i^2$. Note that the set $A$ is closed and bounded ( it is easy to prove it is bounded above by $\sqrt[3]{n-1}$), hence it is compact. Since $f,g$ are continous ( and their first derivatives), then we know that $f$ has global maximum. We have cases that some components lies on boundary or none of them lies. If none of them lies by lagrange method we have $$\frac{1}{|x_1|}=\frac{1}{|x_2|}=...=\frac{1}{|x_n|} \Leftrightarrow x_1^2=x_2^2=...=x_n^2$$If one of the components is zero, then all of them are, whcih gives $f(x)=0 $ which does not help that much, hence assume none of them is $0$ If $n$ is odd, we can't achieve the equality, if $n$ is even, we have $f(x)= nx^2$ If one of the components lies on upper bound, which is $ \sqrt[3]{n-1}$, then other components are $-1$, giving $f(x)= n-1+ \sqrt[3]{n-1}$ If some of the components, say $m$ of them , equals to $-1$, then we can define new functions with degree $n-m$ and reapply things to get that the last components are equal. Then we have equation $$-m+(n-m)y^3=0 \Rightarrow y= \sqrt[3]{\frac{m}{n-m}}$$and we want to maximize $$m+(n-m)y^2= m+ \sqrt[3]{m^2(n-m)}$$Fix $n$, take derivative on $m$ ( real pain), we must have $\frac{(2 n - 3 m) m + 3 ((n - m) m^2)^{2/3}}{3 ((n - m) m^2)^{2/3}}=0$ Overall, before even starting to solve the problem, one may notice the $8/9$ times $ -1$ stuff , hence taking $\boxed{m= \frac{8n}{9}}$, we can see our expression becomes $0$, and it is easy to prove there are no other $0$ root by quadratic calculations, Calculating gives $y=2$. Now, we can check all the values we got and choose the biggest one, which is the latter one, calculating boils down to $\boxed{c= \frac{4}{3}}$, hence we are done!