$l^1$ versus $l^2$

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.

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.