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.



Separable Banach spaces isometric to quotient of a Banach space
The answer is yes. 


Following 


Dowling, P. N.(1-MMOH); Lennard, C. J.(1-PITT-MS)
Every nonreflexive subspace of L1[0,1] fails the fixed point property.  

Proc. Amer. Math. Soc. 125 (1997), no. 2, 443--446,


say that a norm $\|\cdot \|$ on $\ell^1$ is asymptotically isometrically equivalent to the $\ell^1$ norm provided that there exists $\lambda\_n \uparrow 1$ with $\lambda\_1>0$ so that for all sequences $(a\_n)$ of scalars,
$$\sum\_n \lambda\_n |a\_n| \le \| \sum\_n a\_n e\_n \| \le \sum\_n |a\_n|,
$$
where $(e\_n)$ is the usual unit vector basis. Suppose $\| \cdot\|$ satisfies this condition for such a sequence $(\lambda\_n)$. Let $(x\_n)$ be a dense sequence in the unit ball of an arbitrary separable Banach space $X$ and define an operator $Q$ from $(\ell^1, \|\cdot \|)$ to $X$ by mapping $e\_n$ to $\lambda\_n x\_n$ and extending by linearity and continuity. Then $Q$ is a norm one linear operator from $(\ell^1, \|\cdot \|)$ to $X$ such that the image of the unit ball is a dense subset of the unit ball of $X$, and hence $Q$ is a quotient mapping.


Your norm on $\ell^1$ is not is asymptotically isometrically equivalent to the $\ell\_1$ norm. However, look at the closed span $Y$ of $(\sum\_{k\in F\_n} e\_k)\_n$, where $F\_n$ are disjoint finite sets of natural numbers and the cardinalities of $F\_n$ increase to $\infty$. Then $Y$ under your norm is isometric to an asymptotically isometric $\ell^1$ space. Moreover, $Y$ is norm one complemented in your space because the unit vector basis is a symmetric basis in your space, so every subspace spanned by a constant coefficient block basis is contractively complemented.


