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 $\langle P\_{V\_N} e\_i \mid e\_i \rangle = \| P\_{V\_N} e\_i\|\_2^2 \leq 1$.



Thin large subspaces of $\ell^N_1$
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.


