= Solution
Write $K=B_E$ and change coordinates so that its <John ellipsoid> is $B_2^d$, the <Euclidean unit ball>. Then the standard symmetric-body inequalities are
$$
B_2^d\subseteq K\subseteq\sqrt d\,B_2^d,\qquad
\boxed{\|x\|_E\leq |x|\leq\sqrt d\,\|x\|_E}.
$$
We give the contact argument underlying the construction. The <John contact decomposition> provides unit contact vectors $u_j\in\partial K\cap S^{d-1}$ and positive numbers $c_j$ such that
$$
\sum_j c_j u_j\otimes u_j=I,\qquad \sum_jc_j=d,\qquad
|\langle u_j,x\rangle|\leq\|x\|_E.
$$
The last inequality follows because the supporting hyperplane of $K$ at $u_j$ must also support the contained <Euclidean unit ball>, hence has normal $u_j$. For completeness, the decomposition follows from maximal volume as follows. If $I$ were not in the <convex cone> generated by $u\otimes u$ at contact points, finite-dimensional separation would give a <symmetric matrix> $A$ with $\operatorname{tr}A>0$ and $\langle Au,u\rangle\leq0$ at every contact. Subtract $\delta I$ for small $\delta>0$, retaining positive <trace> and obtaining strict negative values. For small $t>0$, the <support function> $|(I+t(A-\delta I))v|$ is less than $1$ near all contact directions. On the compact complement of those neighbourhoods, $h_K(v)-1$ has a positive minimum, so the perturbed <ellipsoid> remains inside $K$ there too. But its <determinant> is $1+t\operatorname{tr}(A-\delta I)+O(t^2)>1$, contradicting maximum volume. The cone is closed: its generating contact matrices form a <compact set> of <trace> $1$, so bounded traces bound their nonnegative coefficient sums. Finite-dimensional <convexity> then supplies finitely many contacts in the decomposition. Finally, for $x\in K$ the identity gives $|x|^2=\sum_jc_j\langle u_j,x\rangle^2\leq d$, proving the second containment above.
Construct an <orthonormal basis> recursively. After $e_1,\ldots,e_{i-1}$ have been chosen, let $H_i$ be their orthogonal complement. The <orthogonal projection> $P_{H_i}$ satisfies
$$
\sum_jc_j|P_{H_i}u_j|^2=\operatorname{tr}P_{H_i}=d-i+1.
$$
As $\sum_jc_j=d$, some contact $v_i$ satisfies $|P_{H_i}v_i|\geq\sqrt{(d-i+1)/d}$. Choose $e_i=P_{H_i}v_i/|P_{H_i}v_i|$. This works through $i=d$ and gives
$$
\|e_i\|_E\geq\langle v_i,e_i\rangle=|P_{H_i}v_i|
\geq\sqrt{\frac{d-i+1}{d}}.
$$
In particular, \b[$\|e_i\|_E\geq1/\sqrt2>1/4$ for $i\leq d/2$].
Keep the contact vectors $v_i$ used for this basis. They lie in $\operatorname{span}(e_1,\ldots,e_i)$. Therefore, if $j<i\leq m=\lfloor d/2\rfloor$, then $\langle v_j,e_i\rangle=0$ and
$$
|v_i-v_j|\geq|\langle v_i,e_i\rangle|\geq1/\sqrt2.
$$
For a <standard Gaussian random vector> $G$ in these coordinates, put $Z_i=\langle v_i,G\rangle$. These centered <Gaussian random variables> have $\mathbb E(Z_i-Z_j)^2=|v_i-v_j|^2\geq1/2$, and $\|G\|_E\geq\max_{i\leq m}Z_i$.
Here is a proof of the <Sudakov-Fernique inequality> needed for this finite family. Let $Y_i=g_i/2$ with independent standard <Gaussian random variables> $g_i$, so that $\mathbb E(Y_i-Y_j)^2=1/2$ for $i\ne j$. Take $Z,Y$ independently and use
$$
F_\beta(z)=\beta^{-1}\log\sum_i e^{\beta z_i},\qquad
p_i(z)=\frac{e^{\beta z_i}}{\sum_j e^{\beta z_j}}.
$$
The Hessian is $\partial_{ij}F_\beta=\beta(p_i\delta_{ij}-p_ip_j)$. Along $Z_s=\sqrt s Z+\sqrt{1-s}Y$, <Gaussian integration by parts> gives
$$
\frac{d}{ds}\mathbb E F_\beta(Z_s)
=\frac\beta4\sum_{i,j}\mathbb E[p_i(Z_s)p_j(Z_s)]
\left(\mathbb E(Z_i-Z_j)^2-\mathbb E(Y_i-Y_j)^2\right)\geq0.
$$
Indeed, the one-dimensional identity $\mathbb E[g\phi(g)]=\mathbb E\phi'(g)$ follows by integrating the standard <Gaussian density> by parts, and applying it to a linear representation of these <Gaussian vectors> gives the covariance formula used here. Since $0\leq F_\beta(z)-\max_i z_i\leq\log(m)/\beta$, letting $\beta\to\infty$ yields $\mathbb E\max Z_i\geq\tfrac12\mathbb E\max g_i$.
To estimate that last maximum, integrating the <Gaussian density> on $[t,t+(t+1)^{-1}]$ gives $\mathbb P(g_1>t)\geq a e^{-t^2/2}/(t+1)$ for an absolute $a>0$. With $t=\tfrac12\sqrt{\log m}$, <independence> shows
$$
\mathbb P(\max_i g_i>t)=1-(1-\mathbb P(g_1>t))^m\longrightarrow1.
$$
Its negative part has expectation at most $\mathbb E[|g_1|\mathbf1_{g_1<0}]\,2^{-(m-1)}$. Hence $\mathbb E\max_i g_i\geq a'\sqrt{\log m}$ for all sufficiently large $m$, and adjusting $a'>0$ covers every $m\geq2$; for these finitely many cases the expectation is strictly positive. For $d\geq4$, $\log\lfloor d/2\rfloor\geq\tfrac13\log d$. For $d=2,3$ use one contact vector and $\mathbb E\|G\|_E\geq\mathbb E|\langle u,G\rangle|=\sqrt{2/\pi}$; for $d=1$ the requested right side is zero. Thus
$$
\boxed{\int\|x\|_E\,d\gamma_d(x)\geq c\sqrt{\log d}}
$$
for an absolute $c>0$, independent of both dimension and the <norm>.
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