We claim that the answer is $0$ when $n$ is even, and $2/n$ otherwise. If $n$ is even, then $x_i = (-1)^i/n$ satisfies $\sum_{i=1}^n |x_i| = 1$ and $\sum_{i=1}^n |x_i+x_{i+1}| = 0$. If $n$ is odd, note that $$|x+y| = \begin{cases} |x| + |y| & x \text{ and } y \text{ have same sign} \\ |x| - |y| & x \text{ and } y \text{ have opposite sign and } |x| \geq |y| \\ |y| - |x| & x \text{ and } y \text{ have opposite sign and } |y| \geq |x|\end{cases}$$so that if we write $|x+y|$ as one of three forms on the right hand side we get at most one minus sign. Writing all terms as the form $|x_i| \pm |x_j|$, we have at most $n-1$ minus sign because $n$ is odd which implies at least one $i$ with $x_i, x_{i+1}$ having the same sign. Furthermore, note that $|x_i|$ with maximum value cannot occur with minus sign. Hence sum of all minus terms (without minus sign) is less than or equal to $\frac{n-1}{n}$ since $\max_{1\leq i\leq n} |x_i| \geq \frac{1}{n}$. Then we have \[ \sum_{i=1}^n |x_i + x_{i+1}| = 2\sum_{i=1}^n |x_i| - 2\cdot (\text{sum of all minus terms}) \geq 2 - 2\cdot \frac{n-1}{n} = \frac{2}{n}. \] Equality occurs when $x_i = (-1)^i/n$ for all $i$.