Thin large subspaces of $\ell^N_1$ Consider a sequence $V_N$ of subspaces of $\ell^N_1$ so that $\dim V_N = N- n$ and $n$ is $\mathsf{o}(N)$. Is it true that these spaces are "thick" (unofficial terminology), i.e. are there constants $K_1,K_2>0$ so that for $N$ large enough,
$$
\sup_{x \in V, \|x\|_1 \leq K_1} \|x\|_\infty \geq K_2
$$
where $\|x\|_\infty$ is (as usual) the supremum of the coordinates of $x$.
A relaxation (or equivalent formulation?) of the above: is there a sequence $x_i \in V_i \subset \ell_1^i \subset \ell_1\mathbb{N}$ which is convergent in the weak$^*$ topology of $\ell_1 \mathbb{N}$ to something else than $0$?
In the $\ell_2$ case, the answer is "overwhelmingly positive". Let $\lbrace e_i\rbrace_{i=1}^N$ be the "standard" basis of $\ell^N_1$. Consider $P_{V_N}$ to be the orthogonal projection on $V_N$, then 
$$
N-n = \dim V_N = \sum_{i=1}^N \langle P_{V_N} e_i \mid e_i \rangle
$$
So that, in fact, there are at least $N-2n$ coordinates $i$ so that  $\langle P_V e_i \mid e_i \rangle \geq \frac{1}{2}$ and $\langle P_{V_N} e_i \mid e_i \rangle = \| P_{V_N} e_i\|_2^2 \leq 1$.
 A: Improved version of my answer.  The following version of Kashin's
(1977) result is needed here: For any $\alpha\in(0,1)$ there
exists $C=C(\alpha)$ such that for any $N$ there is an
$\lceil\alpha N\rceil$-dimensional subspace $L$ of $\ell^N_1$
satisfying
$$\forall x\in L\quad
\frac1{\sqrt{N}}||x||_1\le||x||_2\le\frac{C}{\sqrt{N}}||x||_1$$
(see e.g. Section 1.10.3 in Brazitikos, Giannopoulos, Valettas,
Vritsiou, Geometry of isotropic convex bodies. American
Mathematical Society, Providence, RI, 2014).
This can be used to show that there is a sequence
$(N(i))$ tending to $\infty$ and subspaces $V_{N(i)}$ of  $\ell_1^{N(i)}$ so that dim$V_{N(i)}/N(i)\ge(1-\frac1i)$ and
$$
\sup_{x \in V_{N(i)}, \|x\|_1 \leq 1} \|x\|_2 \le\frac1i.
$$
Since $\|x\|_\infty\le ||x||_2$ this implies that the answer to the first question is `No'.
We apply Kashin's theorem with $\alpha=1-\frac1i$, let $C$ be the
corresponding constant. We pick $N(i)$ so large that $N(i)>i$ and
$$\frac{C}{\sqrt{N(i)}}\le\frac 1i.$$
Let $V_{N(i)}$ be the subspace $L$ of the Kashin's theorem
corresponding to this situation. It is easy to check that all of
the conditions above are satisfied.
A: I discussed this question with Pisier over lunch.  He later called my attention to the paper
[GG] Garnaev, A. Yu.; Gluskin, E. D.;
The widths of a Euclidean ball. (Russian)
Dokl. Akad. Nauk 277 (1984), 1048–1052.
Pisier emailed me,
“They get an equivalent of the relevant  Kolmogorov numbers
of inclusion of $n$-dim Euclidean space into $\ell_\infty^n$.
It says (from my book on the volume p. 81 but there is a misprint- I checked it must be $+1/2$ in the exponent of the log not $-1/2$):
$ L_1^N$ contains a subspace of codimension $N-n$
with Banach-Mazur distance to Hilbert at most
$$
\min \{ \sqrt{N},  (N/n)^{1/2} [\log (1+  N/n)] ^{1/2}\ ”
$$
Thus the OP’s question has a negative answer if $n^{-1}\log (1+  N/n)$ tends to zero; i.e., if  $n^{-1}\log N \to 0$. What if $n\to \infty$ slower than $\log N$? 
Take an $n$ codimensional subspace $V_N$ of $\ell_1^N$.  Take an Auerbach basis $y_1,\dots, y_n$ for $V_N^\perp \subset \ell_\infty^N$.
Let $1/\epsilon $ be a positive integer; something like $\epsilon = 1/10$ is OK. Tile $[-1, 1]^n$ with boxes of side length $\epsilon/n$; the number of boxes is around $(2n\epsilon/n)^n$. Consequently, if $N> (2n\epsilon)^n$, there exist $1\le i < j \le N$ s.t. $(\langle y_k, e_i \rangle)_{k=1}^n$ and $(\langle y_k, e_j \rangle)_{k=1}^n$ lie in the same box.  Since $y_1,\dots, y_n$ is Auerbach, this means that 
$|\langle y, e_i -e_j\rangle| < \epsilon $ for all norm one $y \in V_N^\perp$.  That is, the distance of $e_i -e_j$ to $V_N$ is less than $\epsilon$. Since $\|e_i -e_j\|_1 = 2$ and $\|e_i -e_j\|_\infty =1$, this gives a positive answer to the OP’s question if $N > (2n\epsilon)^n$ for infinitely many $N$; i.e., if $n\log n < \delta \log N$ for a computable $\delta$.  
I don’t know where the break point occurs.
A: By Kashin's theorem you can decompose $\ell_1^{2n}$ into two orthogonal subspaces each of dimension $n$ so that you have $\|x\|_1$ is about $\sqrt{n}\|x\|_2$ for all $x$ in each of these subspaces. So by a trivial estimate between $\|x\|_2$ and $\|x\|_{\infty}$ norms you get $\|x\|_{\infty}$ less than in the order of $1/\sqrt{n}$.
