Let $T:(\mathbb R^n,\|.\|_1)\to (\mathbb R^n,\|.\|_2)$ be a linear isomorphism. Then clearly it is a topological isomorphism also.
I want to show that $\|T\|\|T^{-1}\|\geq \sqrt{n}$.
Here $\|(x_1,\ldots,x_n)\|_1=\sum\limits_{i=1}^n |x_i|$ and $\|(x_1,\ldots,x_n)\|_2=(\sum\limits_{i=1}^n |x_i|^2)^{\frac{1}{2}}$. I feel the inequality $\|x\|_1\leq \sqrt{n}\|x\|_2$ will be somehow used. But I could not show it. Any help will be appreciated.
On
i) Assume that $\| T\|=1$ (In fact, we are sufficient : If $\| S\|=C$, then let $T=\frac{1}{C}S$ so that $\|T\|=1$. And $ \| S\|\|S^{-1}\|>t $ iff $\|T\|\|T^{-1}\|>t$)
Assume that $\|Te_i\|_2=1$ where $e_i$ is canonical basis.
ii) If $ \|v\|_1=1$, then $\|Tv\|_2\leq 1$ so that $\|\ \|_1$-unit sphere $S_1$ is mapped into $\|\ \|_2$-unit sphere $S_2$ .
We define a center of face of $S_1$ : $$x= \frac{1}{n}\sum_i\ \varepsilon_ie_i $$ where $\varepsilon_i\in \{\pm 1\}$. Then from iii), there is $x$ with $$ \|Tx\|_2\leq \frac{1}{\sqrt{n}} \ \ast$$
If so, $$ \bigg\|T^{-1} \frac{Tx}{\| Tx\|_2 } \bigg\|_1 \geq \sqrt{n} $$
iii) Assume that $\{v_i\}_{i=1}^n,\ \|v_i\|_2=1$ is independent. Prove that $$A:=\bigg\|\frac{1}{n}\sum_{i=1}^n \ \varepsilon_i v_i \bigg\|_2 \leq \frac{1}{\sqrt{n}} $$ for some $\varepsilon_i\in \{ \pm 1\}$.
Proof : If $w_i=\varepsilon_iv_i$ and $n=2$, then there is an obtuse triangle $w_1w_2o$ where $o$ it the origin. Hence $A=\cos\ \theta <\frac{1}{\sqrt{2}},\ \frac{\pi}{4}<\theta$.
We will use an induction. Assume that $\bigg\| W:= \frac{\sum_{i=1}^{n-1}\ w_i}{n-1} \bigg\|_2\leq \frac{1}{\sqrt{n-1}}$
Assume that $$ w_n=w_\perp +w $$ is orthogonal decomposition where $w_\perp \perp ( w_1,\cdots, w_{n-1})$.
Then \begin{align*}&\ \ \ \bigg\| \frac{ (n-1)W +w_\perp +w }{n}\bigg\|_2^2 \\&=\frac{ \| w_\perp\|_2^2 }{n^2} + \frac{ \|w\|_2^2 + (n-1)^2\|W\|_2^2}{n^2} + 2\ \frac{n-1}{n}\ W\cdot w \\& \leq 2\ \frac{n-1}{n}\ W\cdot w + \frac{1}{n} \end{align*}
Since we can use $-w_n$, then the proof is followed.
Remark : In the above, it holds for $ \|v_i\|_2\leq 1$.
EDIT: THIS ANSATZ IS WRONG. SEE BELOW FOR AN ALTERNATIV ANSATZ.
I don't think that this inequality is true for $n \geq 2$.
Let $T := \operatorname{id}_{\mathbb{R}^n}$, i.e. $$T \colon (\mathbb{R}^n, \|\cdot\|_1) \to (\mathbb{R}^n, \|\cdot\|_2), \quad x \mapsto x.$$ This an isomorphism of normed spaces with $T = T^{-1}$ and $\| T \| = 1$.
I think this ansatz should do the trick [EDIT: NO IT DOESN'T]: $$ \sqrt{n} = \| \operatorname{id}_{\mathbb{R}^n} \|_{2,1} = \| T T^{-1} \|_{2,1} \leq \|T \|_{2,2} \|T^{-1}\|_{2,1} \leq \|T \|_{1,2} \|T^{-1}\|_{2,1}. $$
In the last inequality I used that $$ \|T\|_{2,2} \leq \|T\|_{1,2}, $$ as for a matrix representation $T = (a_{ij})_{i,j=1}^n$ we have that $$ \|T\|_{2,2} = \Big( \sum_{i,j = 1}^n a_{ij}^2 \Big)^{1/2} = \Big( \sum_{j = 1}^n \| a_{.j} \|_2^2 \Big)^{1/2} \leq \Big(\sum_{j = 1}^n \| a_{.j}\|_1^2 \Big)^{1/2} = \|T\|_{1,2} $$
Apparently the relation between operator norms and entry-wise norms is particularly difficult (in fact it is NP-hard), see here and here.