$p_i, i\leq n$ span the space of all polynomials of degree at most $n$. If $q$ is any monic polynomial of degree $n$ then $q-p_n$ is orthogonal to $p_n$ because degree of $q-p_n$ is at most $n-1$ and hence it belongs to the span of $p_i:i<n$. Hence $\|q\|^{2}=\|(q-p_n)+p_n\|^{2}=\|q-p_n\|^{2}+\|p_n\|^{2}$ which implies $\|q\|^{2} \geq \|p_n\|^{2}$.
Yes, they form a basis of the space of even degree polynomial. Even more generally, if $(P_n)_{n \ge 0}$ is a basis of the space of polynomials, then $\left(P_n\left(\frac{X^2}{2}-1 \right)\right)_{n \ge 0}$ is a basis of the space of even degree polynomial :
To prove this, denote $E$ the space of odd degree polynomials, and observe that the linear map
$$\begin{matrix}
f : & \mathbb{R}[X] & \longrightarrow & E\\
& P(X) & \longmapsto & P \left(\frac{X^2}{2}-1 \right)\end{matrix}$$
is an isomorphism and hence it maps a basis of $\mathbb{R}[X]$ to a basis of $E$.
Indeed, let's prove this is indeed an isomorphism. First observe that $f$ does indeed map into $E$ because $P \left(\frac{X^2}{2}-1 \right)$ is always even for all $P$.
Injectivity : if $f(P) = 0$, then we have $\forall x \in \mathbb{R}, P \left(\frac{x^2}{2}-1 \right) = 0$, which implies that $P$ is zero on $[-1,\infty[$ so $P = 0$.
Surjectivity : let $Q(X) \in E$ be an even degree polynomial. We can write it as $Q(X) = \sum_{k = 0}^{s} a_{2k} X^{2k}$. So by setting $R(X) = \sum_{k = 0}^{s} a_{2k} X^{k} \in \mathbb{R}[X]$, we get $Q(X) = R\left(X^2\right)$. And thus by setting $P(X) = R(2(X+1)) \in \mathbb{R}[X]$, we get $R(X) = P\left(\frac{X}{2} - 1 \right)$ so
$$Q(X) = R\left(X^2\right) = P\left(\frac{X^2}{2} - 1 \right)$$
which proves that $Q = f(P)$.
Best Answer
It is standard $L^2$ on $[-1,1]$, and the orthogonal polynomials are the Legendre polynomials. Or you can calculate the distance from your function to the space $\langle 1, x, \ldots, x^n\rangle$ using Gram determinants.
$\bf{Added:}$
Let us do some calculations. The Legendre $P_n(x)$ polynomials $P_n(x)$ on $[-1,1]$ are an orthogonal system. $P_n(x)$ is of degree $n$, and orthogonal to $1$, $x$, $\ldots$, $x^{n-1}$. This leaves it uniquely determined up to a normalizing constant, which it is such that $P_x(1)=1$. Here is a list of the first $6$ polynomials.
$$ 1, x, \frac{1}{2}(-1 + 3 x^2), \frac{1}{2}(-3 x + 5 x^3), \frac{1}{8}(3 - 30 x^2 + 35 x^4), \frac{1}{8}(15 x - 70 x^3 + 63 x^5)$$
The projection of a vector on a subspace spanned by several orthogonal vectors $w_k$ is $$\pi(v) = \sum_k \frac{\langle v, v_k\rangle v_k}{\langle v_k, v_k\rangle}$$ and the distance squared from $v$ to this subspace is $$d^2 =\|v\|^2 - \sum_k \frac{|\langle v, w_k\rangle|^2}{\|w_k\|^2}$$
Now, for the Legendre polynomials we have $\|P_n(x)\|^2 = \frac{2}{2n+1}$.
Another approach is with Gram matrices and determinants. The distance squared from $x^{\alpha}$ to a space of functions $\langle x^{\alpha_i} \rangle$ equals $G(x^{\alpha}, x^{\alpha_i})/ G( x^{\alpha_i})$.
Notice that $x^{2/3}$ is even, so it will project into even functions. Recall that for $\beta$ even we have $\int_{-1}^1 x^{\beta} = 2/(\beta+1)$. Therefore, for $\alpha_i$ even the Gram determinant $G(x^{\alpha_i})$ is a Cauchy dererminant $$\det(\frac{2}{\alpha_i + \alpha_j+1}) = 2^{|I|} V(\alpha_i)^2 \cdot \prod_{i,j} \frac{1}{\alpha_i + \alpha_j + 1}$$ where $V$ is a Vandermonde determinant.
We conclude that the distance squared from even $x^{\alpha}$ to $\langle 1, x^2, \ldots, x^{2n}\rangle$ equals $$d^2 = \frac{2}{2\alpha+ 1} \prod_{k=0}^n \frac{(2k -\alpha)^2}{(2k +\alpha + 1)^2}$$