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.