= Solution
Projection onto the constant <polynomials> requires the residual to be <orthogonal> to $1$. For the original <inner product>,
$$
0=\int_{-1}^1(x^2-p_0)\,dx=\frac23-2p_0,\qquad \boxed{p_0=\frac13.}
$$
For the weighted <inner product> the corresponding <normal equation> is
$$
0=\int_{-1}^1x^{2n}(x^2-\bar p_0(n))\,dx=\frac2{2n+3}-\bar p_0(n)\frac2{2n+1}.
$$
Hence
$$
\boxed{\bar p_0(n)=\frac{2n+1}{2n+3}=1-\frac2{2n+3}\longrightarrow1.}
$$
It increases monotonically towards one. Although the weight vanishes at zero, it is positive elsewhere, and a nonzero continuous function is nonzero on an interval of positive weight. Thus the weighted expression really is an <inner product> and the constant minimizer is unique.
The normalized weight $(2n+1)x^{2n}/2$ has total mass one and assigns mass $r^{2n+1}$ to $[-r,r]$ for each $0<r<1$. Its mass therefore concentrates near $-1$ and $1$, where $x^2=1$. This explains why the weighted best constant tends to one rather than to the unweighted spatial average. It is <endpoint concentration in weighted constant approximation>.
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