You can use Lagrange multipliers. There are probably good books that do a better job explaining this subject than this Wikipedia article.
EDIT: What was earlier a conjecture is now proved.
Following this approach, the maximal value is one half of the largest eigenvalue of a certain matrix $M$ below.
You have a function to optimize: $$F(\vec{x})=\sum x_ix_{i+1}$$
subject to two constraints:
\begin{align}G(\vec{x})&=\sum x_i=0\\
H(\vec{x})&=\sum x_i^2=1
\end{align}
With such simple polynomial functions as these, the method of Lagrange multipliers states that if $\vec{x}$ is a potential extremal point, then for some $\lambda$ and $\mu$,
\begin{align}\nabla F&=\lambda\nabla G+\mu\nabla H\\
M\vec{x} & = \lambda\vec{1}+2\mu\vec{x}
\end{align}
where
\begin{align}
M&=\
\begin{bmatrix}0&1&0&0&\cdots&0&1\\
1&0&1&0&\cdots&0&0\\
0&1&0&1&\cdots&0&0\\
0&0&1&0&\cdots&0&0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots&\vdots\\
0&0&0&0&\cdots&0&1\\
1&0&0&0&\cdots&1&0\end{bmatrix}\\
\vec{1}&=\begin{bmatrix}1\\1\\1\\1\\\vdots\\1\\1\end{bmatrix}
\end{align}
Summing the equation $M\vec{x} = \lambda\vec{1}+2\mu\vec{x}$ over all rows and using the first constraint shows than $\lambda=0$, and $\vec{x}$ must be an eigenvector of $M$ in the eigenspace $V_{2\mu}$. This gives more constraints: $J(\vec{x})=(M-2\mu I)\vec{x}=0$. Since $\nabla F$ is a linear combination of $\nabla J$ and $\nabla H$, $F$ must be constant subject to the constraints $H(\vec{x})=1$ and $J(\vec{x})=0$.
So $F$ takes constant values on the eigenspaces of $M$ intersected with the sphere given by $H(\vec{x})=1$. "All" that remains is to find one eigenvector from each eigenspace of $M$ (other than $V_2$ which is orthogonal to the constraint $G(\vec{x})=0$) and compute $F$. I do not know a way to handle this matrix $M$ for all values of $n$ simultaneously though.
If $\vec{x}$ is an eigenvector for $M$ with eigenvalue $2\mu$ satisfying $H(\vec{x})=1$, then
\begin{align}
F(\vec{x}) & =\vec{x}^t\left(\frac{1}{2}M\right)\vec{x}\\
&=\vec{x}^t\frac{1}{2}(2\mu\vec{x})\\
&=\mu\ \vec{x}^t\vec{x}\\
&=\mu
\end{align}
So in summary, the only potential extremal points for $F$ happen at the intersections of the unit sphere with the various eigenspaces of $M$. In these intersections, $F$ has constant value $\mu$, which is one half of the eigenvalue of $M$ for that eigenspace. If you can find the eigenvalues of $M$, you have the answers to your question.