Inner products and norms

I know that every inner product induces a norm (that induces a metric), and a norm is induced by an inner product if it satisfies the parallelogram law. I know that $$x^Tx$$ induces $$\begin{Vmatrix} x \end{Vmatrix}_2^2$$ by definition of norm $$\begin{Vmatrix} x \end{Vmatrix}:=\sqrt{x^Tx}$$, and at the same time $$\begin{Vmatrix} x \end{Vmatrix}_2^2$$ is the norm induced by inner product $$x^Tx$$. I also know that $$\begin{Vmatrix} x \end{Vmatrix}_1^2$$ and $$\begin{Vmatrix} x \end{Vmatrix}_{\infty}^2$$ are not induced by inner products because it's not true that $$\forall x,y\in \mathbb{R}^n$$

$$\begin{Vmatrix} x+y \end{Vmatrix}_{1,\infty}^2+\begin{Vmatrix} x-y \end{Vmatrix}_{1,\infty}^2= 2(\begin{Vmatrix} x \end{Vmatrix}_{1,\infty}^2+\begin{Vmatrix} y \end{Vmatrix}_{1,\infty}^2).$$

Then, starting from $$x^TAx$$ with $$A=LL^T$$ the Cholesky decomposition of $$A$$, is there a way to get $$\begin{Vmatrix} x \end{Vmatrix}_{1}^2$$ and $$\begin{Vmatrix} x \end{Vmatrix}_{\infty}^2$$? Can I still write $$\begin{Vmatrix} x \end{Vmatrix}_{1}^2=x^Tx$$ or $$\begin{Vmatrix} x \end{Vmatrix}_{\infty}^2=x^Tx$$ even if these norms are not induced by $$x^Tx$$?

No. No matter which bilinear form $$(x,y)\mapsto x^T A y$$ you choose, a norm not satisfying the parallelogram identity can never be induced as $$\|x\|=\sqrt{x^T A x}$$.