Assume that $p = 3$. Then obviously $x_1$ can take any value. Now suppose that $p \ge 5$. We want to prove that $(x_1, x_2, \dots, x_{\frac{p-1}{2}}) \in \{0, 1\}^{\frac{p-1}{2}}$. Note that
$$\sum_{i=1}^{\frac{p-1}{2}} (1 - a x_i)^{\frac{p-1}{2}} \equiv \frac{p-1}{2} + M \left( (1-a)^{\frac{p-1}{2}} - 1 \right) \pmod{p},$$
where $M$ is common value mod $p$ of
$$\sum_{i=1}^{\frac{p-1}{2}} x_i^j \quad \left(1 \le j \le \frac{p-1}{2}\right).$$
If we choose $a$ such that $(1-a)^{\frac{p-1}{2}} = 1$, then by using $(1-ax_i)^{\frac{p-1}{2}} = \pm 1$ or $0$ and the above equation we have for each $i$,
$$(1 - a x_i)^{\frac{p-1}{2}} = \left( \frac{1 - a x_k}{p} \right) = 1$$
So if $\left(\frac{t}{p}\right) = 1$ we get
$$\left( \frac{1 - (1 - t)x_i}{p} \right) = 1$$
Now if $x_i \ne 0$ then the map $f(t) = 1 - (1-t)x_i$ is bijective. Hence if $P$ is the set of square remainder, $f$ define a bijection from $P$ to itself and
$$\sum_{t \in P} f(t) = \sum_{t \in P} t = 0,$$
which implies that $x_i = 1$.