I want to prove that if the interval $[a,b]$ is symmetric (respect to origin) and if $w(x)$ (the weight function) is an even function, then the Gaussian nodes will be symmetric and $A_i=A_{n-i}$ where; $$A_i=\int_a^b l_i(x)w(x)$$
and
$$l_i(x)= \Pi_{j=0, \ j\neq i}^n \frac{x-x_j}{x_i-x_j}$$
I will use a bit different notation. I consider the nodes $x_1,\dots,x_n$ instead of $x_0,\dots,x_n$ as you proposed. I feel better with my notation. Here $$l_i(x)=\prod_{j=1,j\ne i}^n\frac{x-x_j}{x_i-x_j}.$$ Then we need to show that $x_i=-x_{n-i+1}$ and $A_i=A_{n-i+1}$ for $i=1,\dots,n$.
Let $P_n$ be the $n$-th degree orthogonal polynomial on $[-a,a]$ wrt. $w$. Then $P_n$ is orthogonal to each polynomial $p$ of degree $\le n-1$, so (because $p(-x)$ is also a polynomial of degree $\le n-1$), $$\int_{-a}^a P_n(x)p(-x)w(x)dx=0.$$ By substitution ($-x$ instead of $x$) and since $w$ is even, we arrive at $$0=\int_{-a}^a P_n(x)p(-x)w(x)dx=\int_{-a}^a P_n(-x)p(x)w(-x)dx=\int_{-a}^a P_n(-x)p(x)w(x)dx.$$ Hence $Q(x)=P_n(-x)$ is orthogonal to each polynomial of degree $\le n-1$, then $Q$ is also an orthogonal polynomial of $n$-th degree. So, both $P$ and $Q$ have the same roots, from which the symmetry of nodes of Gaussian quadrature follows.
To show that the coefficients are symmetric, try to prove that $l_{n-i+1}(x)=l_i(-x)$. I leave this as an exercise for you (of course, you should apply the symmetry of nodes, which I demonstrated above). The hint could be another form of $l_i$: $$l_i(x)=\frac{P_n(x)}{(x-x_i)P_n'(x_i)}$$ together with the fact that $P_n$ is an even function for even $n$ and odd function for odd $n$.
Next integrate these polynomials by the same trick with even weight function:
\begin{multline*} A_{n-i+1}=\int_{-a}^a l_{n-i+1}(x)w(x)dx=\int_{-a}^a l_{n-i+1}(-x)w(-x)dx\\ =\int_{-a}^a l_{n-i+1}(-x)w(x)dx=\int_{-a}^a l_i(x)w(x)dx=A_i. \end{multline*}