We know that every separable Banach space is isometrically isomorphic to a quotient space of $(\ell^1,\|.\|_1)$. We also know that the norm defined by $\|x\|=(\|x\|_1^2+\|x\|_2^2)^{1/2}$ for all $x\in \ell^1$ is equivalent to $\|.\|_1$. My question is that is every separable Banach space isometrically isomorphic to a quotient of $(\ell^1,\|.\|)$?

The standard way of proving the result that I stated above is as follows:

Let $\{x_n:n\in \mathbb{N}\}$ be a dense subset of $S_X$, where $X$ is a separable Banach space. Then $$T((\lambda_n))=\sum\limits_{n=1}^{\infty}\lambda_n x_n \text{ for all }(\lambda_n)\in \ell^1,$$ is a continuous linear map from $\ell^1$ onto $X$. Consequently, $\ell^{1}/\ker T$ is linearly homeomorphic to $X$. It can also be shown that $\ell^1/\ker T$ and $X$ are actually isometric. I tried to mimic the same proof for my question too, but couldn't succed. Any help is appreciated.