Is this norm induced by an inner product? Is $||x||=\sqrt{\sum_{n=2}^{\infty} \frac{1}{n} x^2_n} +|x_1|$ induced by an inner product?
I claim that this is not true because in a previous question I found that the inner product $<x,y>= \sum_{n=2}^\infty \frac{1}{n}x_ny_n+x_1y_1$ deduces the norm $||x||=\sqrt{<x,y>}$.
What specific $x_n$-sequence makes the parallelogram fail?

The obvious candidates are the vectors $x=(1,0,0,\ldots) $ and $y=(0,1,0,0,\ldots) $. You have
$$
\|x+y\|^2+\|x-y\|^2=2(1+\sqrt{1/2})^2\ne 2 (1+1/2)=2\|x\|^2+2\|y\|^2.
$$
I cannot make sense of your second paragraph. 
