Check this is a hilbert norm: $ \ell^2 $ with norm $\| \cdot \| := \| \cdot \|_{\ell^2} + \| \cdot \|_{\ell^p}$ Clearly $ p \geq 2 $ so it gains sense calculating the $\ell^p $-norm.
According to my calculation this norm is equivalent to the $\ell^2 $ norm, in fact given a cauchy sequence w.r.t $\| \cdot \| $ it his a cauchy sequence w.r.t. each norm. Using the completeness of $\ell^2 $ and the fact that $\|u\|_{\ell^p}^p \leq \| u \|_{\ell^2}^2 $ i'm forcing the the sequence to converge to the limit of the norm 2. So it is complete, and by the inverse mapping theorem $\ell^2$ norm is equivalent to my norm (used the identity map).
So this norm is hilbertizable. But i can't prove in an efficient way that is or (isn't) an hilbert norm

In order for a norm, of a normed space to come from an inner product, it has to satisfy the parallelogram identity:
$$
\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2.
$$
Your norm does not satisfy such an identity, i.e.,
$$
x=e_1,\,\,y=e_2,\quad \|x\|=\|x\|_2+\|x\|_p=2=\|y\|,\,\, 
$$
$$
\|x-y\|=\|x+y\|=\sqrt{2}+\sqrt[p]{2}
$$
Thus 
$$
\|x+y\|^2+\|x-y\|^2=(2\sqrt{2}+2\sqrt[p]{2})^2,
$$
while
$$
2\|x\|^2+2\|y\|^2=8.
$$
