$l^1$ versus $l^2$ Is there an elementary proof of this Banach space fact?

If the Banach space $V$ is linearly isomorphic to $l^1$, then it does not isometrically contain euclidean spaces of arbitrarily large finite dimension, i.e., a copy of $l^2_n$ for all $n$.

Failing that, good references on the subject?
 A: Is it true? $l^1$ is a sum of finite-dimensional $l_n^1$ over $n=1,2,\dots$. In summands you have almost spherical sections of large dimensions by Dvoretzky theorem, this allows to change norm a bit so that unit balls in summands contain large spherical sections. 
A: There does not exist an elementary proof. Sadly, there is no proof at all, because the fact is false. Here is a construction of an $\ell^1$-space containing every $\ell^2_n$.
Consider the normalized Gaussian measure $d\mu=\pi^{-1/2}e^{-t^2}dt$. Form its infinite dimensional tensor product $d\mu_\infty$ over ${\mathbb R}^{\mathbb N}$. Now let $E$ denote the vector space of linear functions
$$f_a:x\longmapsto a\cdot x,\qquad x\in{\mathbb R}^{\mathbb N},$$
where $a$ has finite support. $E$ is contained in $L^1(d\mu_\infty)$. One sees easily, from the rotational invariance of $d\mu$, that
$$\|f_a\|_1\left(=\int|f_a(x)|d\mu_\infty(x)\right)=C\|a\|_2$$
where
$$C=\int_{\mathbb R}|t|d\mu(t).$$
Hence $L^1(d\mu_\infty)$ contains every $\ell^2_n$.
Nota. This argument is used to prove that Euclidian spaces satisfy the Hlawka Inequality. The latter is almost trivial in an $\ell^1$ space.
