= Solution
For a $d$-dimensional <normed vector space>, with <unit ball> $K$ and <John ellipsoid> $\mathcal E$, its <volume ratio> is
$$
\operatorname{vr}(E)=\left(\frac{\operatorname{vol}K}{\operatorname{vol}\mathcal E}\right)^{1/d}.
$$
It is at least $1$ and invariant under invertible linear coordinate changes.
For $\ell_1^n$, the <unit ball> has volume $2^n/n!$: its intersection with each orthant is the standard simplex of volume $1/n!$. Its <John ellipsoid> is $n^{-1/2}B_2^n$. To verify maximum volume, let $AB_2^n$ be any centered inscribed <ellipsoid>. For each sign vector $\varepsilon\in\{-1,1\}^n$, containment gives $|A^T\varepsilon|\leq1$. Averaging over independent <Rademacher random variables> yields $\operatorname{tr}(AA^T)\leq1$. The <arithmetic-geometric mean inequality> for its squared <singular values> gives $|\det A|\leq n^{-n/2}$. The scalar <matrix> $n^{-1/2}I$ attains this bound because $\|x\|_1\leq\sqrt n|x|$. Centered <ellipsoids> suffice: if $a+AB_2^n$ is contained in the symmetric convex <unit ball>, averaging it with $-a+AB_2^n$ gives the centered copy with the same volume.
Using the stated <Euclidean ball> volume and $n=2k$,
$$
\operatorname{vr}(\ell_1^{2k})^{2k}
=\frac{2^{2k}(2k)^k k!}{(2k)!\pi^k}
\leq\left(\frac8\pi\right)^k,
$$
since $(2k)!/k!=\prod_{j=k+1}^{2k}j\geq k^k$. Therefore \b[$\operatorname{vr}(\ell_1^{2k})\leq\sqrt{8/\pi}$ uniformly in $k$].
\b[The subsequent assertion for an arbitrary $2k$-dimensional space is false as printed.] Here is a counterexample independent of any asymptotic section theorem. Take $E=\ell_\infty^{2k}$. Its <John ellipsoid> is $B_2^{2k}$: an inscribed $AB_2^{2k}$ has every row of $A$ of Euclidean length at most $1$, so the <Hadamard inequality> gives $|\det A|\leq1$, attained by $A=I$. Let $F$ be any $k$-dimensional subspace and $f_1,\ldots,f_k$ an <orthonormal basis> of it. The random unit vector
$$
x=\frac1{\sqrt k}\sum_{j=1}^k\varepsilon_j f_j
$$
has $|x|=1$. Since $\sum_j f_j(i)^2\leq1$ for each coordinate $i$, the <moment-generating function> bound for independent <Rademacher random variables> gives
$$
\mathbb E e^{t x_i}\leq e^{t^2/(2k)},\qquad
\mathbb P(\|x\|_\infty>r)\leq4k\,e^{-kr^2/2}.
$$
Choose $r=\sqrt{2\log(8k)/k}$; the latter <probability> is at most $1/2$, so some unit vector in $F$ has $\|x\|_\infty\leq r$. The proposed section inequality would therefore force
$$
\boxed{L\geq\sqrt{\frac{k}{2\log(8k)}}\longrightarrow\infty}.
$$
Thus even one half-dimensional section cannot have the printed universal constant.
The valid <orthogonal splitting with bounded volume ratio> is that $L$ may depend on $V=\operatorname{vr}(E)$, and \b[$L=32V^2$ suffices]. This supplies a dimension-independent constant for the preceding $\ell_1^{2k}$ case, or whenever $V$ is bounded uniformly. We prove the complete qualified assertion, including its <epsilon-net> ingredient.
In <John position>, $\|x\|_E\leq|x|$. Polar integration of the radial function $\rho(u)=1/\|u\|_E$ gives
$$
V^{2k}=\frac{\operatorname{vol}K}{\operatorname{vol}B_2^{2k}}
=\int_{S^{2k-1}}\|u\|_E^{-2k}\,d\sigma(u),\qquad
\sigma\{\|u\|_E\leq r\}\leq(Vr)^{2k},
$$
where $\sigma$ is normalized spherical measure.
A maximal $\varepsilon$-separated subset of $S^{k-1}$ is an <epsilon-net>. Balls of radius $\varepsilon/2$ about its points have disjoint interiors and lie in the ball of radius $1+\varepsilon/2$. Comparing volumes bounds its <cardinality> by $(1+2/\varepsilon)^k$. Finite existence follows from this packing bound, and maximality proves the net property without assuming a separate covering theorem.
Choose two orthogonal $k$-dimensional subspaces $F_0,F_0^\perp$ and one such net on each <unit sphere>. Rotate both by a common uniformly distributed orthogonal <matrix> $U$. This distribution can be obtained from Question 1 applied to the compact <orthogonal group> with its Frobenius <metric>; left and right multiplication are transitive <isometries>. Consequently $Uv$ is uniformly distributed on $S^{2k-1}$ for every fixed unit vector $v$. The <union bound> shows that the <probability> that some rotated net point has <norm> less than $2\varepsilon$ is at most
$$
2(1+2/\varepsilon)^k(2V\varepsilon)^{2k}
=2\left[4V^2(\varepsilon^2+2\varepsilon)\right]^k.
$$
Set $\varepsilon=(32V^2)^{-1}$. Because $V\geq1$, the bracket is at most $1/4+1/256$, and twice its $k$th power is less than $1$ for every $k\geq1$. Some rotation therefore makes every point of both nets have <norm> at least $2\varepsilon$.
The <triangle inequality> and $\|x\|_E\leq|x|$ imply that $\|\cdot\|_E$ is $1$-<Lipschitz> for the <Euclidean metric>. Every unit vector of $UF_0$ or $UF_0^\perp$ is within $\varepsilon$ of a good net point and thus has <norm> at least $\varepsilon$. By homogeneity, on both orthogonal subspaces,
$$
\boxed{\|x\|_E\leq |x|\leq32\operatorname{vr}(E)^2\|x\|_E}.
$$
For $\ell_1^{2k}$ the first computation gives the absolute choice $L=256/\pi$. The missing bounded-<volume ratio> restriction is essential; the literal unrestricted assertion cannot be proved.
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