= Solution
For a <bounded linear operator> $T:E\to F$, write the weak square norm of a finite family as
$$
w_2(x_1,\ldots,x_m)=\sup_{\phi\in E^*,\ \|\phi\|\le1}\left(\sum_{j=1}^m|\phi(x_j)|^2\right)^{1/2}.
$$
It is a <2-summing operator> if $(\sum_j\|Tx_j\|^2)^{1/2}\le Cw_2(x_1,\ldots,x_m)$ for all finite families, and $\pi_2(T)$ is the least such $C$. This specializes the definition of an <absolutely p-summing operator>.
First let $E=H_1$, $F=H_2$ be <Hilbert spaces>. For the column operator $A:\ell_2^m\to H_1$, $Ae_j=x_j$, the <Riesz representation theorem> gives $w_2(x_1,\ldots,x_m)=\|A^*\|=\|A\|$. If $T$ is a <Hilbert-Schmidt operator>, the <Parseval identity> applied to orthonormal bases gives
$$
\sum_{j=1}^m\|Tx_j\|^2=\|TA\|_{\mathrm{HS}}^2
=\sum_\alpha\|A^*T^*f_\alpha\|^2
\le\|A\|^2\sum_\alpha\|T^*f_\alpha\|^2
=\|A\|^2\|T\|_{\mathrm{HS}}^2.
$$
Thus $\pi_2(T)\le\|T\|_{\mathrm{HS}}$. Conversely test the summing inequality on any finite orthonormal family in $H_1$. Its weak square norm is $1$, by the <Bessel inequality>, with equality by choosing one of its vectors. Hence $\sum_j\|Te_j\|^2\le\pi_2(T)^2$. Taking the supremum over finite subsets of an <orthonormal basis> proves the <Hilbert-Schmidt operator> condition and the reverse norm inequality. These basis sums work even for nonseparable <Hilbert spaces>, being understood as suprema of finite subsums. Therefore
$$
\boxed{T\text{ is Hilbert-Schmidt}\iff T\text{ is 2-summing},\qquad \pi_2(T)=\|T\|_{\mathrm{HS}}.}
$$
For an $n$-dimensional <normed vector space> $E$, again associate a family $(x_j)$ with $A:\ell_2^m\to E$. By the <Hahn-Banach theorem> and scalar Euclidean duality,
$$
\|A\|=\sup_{\|\phi\|\le1}\left(\sum_j|\phi(x_j)|^2\right)^{1/2}=w_2(x_1,\ldots,x_m).
$$
Let $Q$ be the <orthogonal projection> onto $(\ker A)^\perp$, whose dimension is at most $n$. Since $A=AQ$, the <trace> of $Q$ gives
$$
\sum_j\|x_j\|_E^2\le\|A\|^2\sum_j\|Qe_j\|_2^2
=\|A\|^2\operatorname{rank}A\le n\,w_2(x_1,\ldots,x_m)^2.
$$
This proves $\pi_2(I_E)\le\sqrt n$ for every norm, without comparing that norm to an arbitrary Euclidean norm.
To obtain equality and the projection bound, choose a maximum-volume centered <ellipsoid> inside the unit ball $B_E$ and use coordinates in which it is the Euclidean ball $B_2$. In the real case this is the <John ellipsoid>; in the complex case maximize among Hermitian ellipsoids. There are unit contact vectors $u_r\in\partial B_E\cap\partial B_2$ and positive weights $c_r$ with the <John contact decomposition>
$$
\sum_r c_r u_ru_r^*=I_E,\qquad \sum_r c_r=n.
$$
For completeness, the maximum-volume argument gives this identity. If $I_E$ were outside the closed convex cone of the contact matrices, finite-dimensional separation would supply a symmetric, or Hermitian, matrix $D$ with $\operatorname{tr}D>0$ and $u^*Du\le0$ at every contact. The cone is closed because its compact generators have trace one. Subtract a sufficiently small multiple of the identity to make all contact inequalities strict while retaining positive trace. The support function of $(I+tD)B_2$ decreases near the contact directions for small $t>0$; away from them the strict gap between the support functions of $B_E$ and $B_2$ permits the same small deformation. Thus this deformed ball still lies in $B_E$. Its volume increases, since the real determinant is $1+t\operatorname{tr}D+O(t^2)$ in the real case and $|\det_{\mathbb C}(I+tD)|^2=1+2t\operatorname{tr}D+O(t^2)$ in the complex case. This contradicts maximality. The finite-dimensional convex-cone representation gives finitely many contacts, and taking traces gives the sum of the weights.
At a contact $u_r$, a supporting hyperplane to $B_E$ also supports $B_2$. The Euclidean ball has a unique supporting direction there, so the corresponding functional is $\phi_r(x)=u_r^*x$ and has dual norm one on $E$. In the complex case balancedness of $B_E$ turns the real support bound into $|\phi_r(x)|\le\|x\|_E$. Also $B_2\subseteq B_E$ gives $\|x\|_E\le|x|_2$. For any dual-unit functional $\phi$, this containment implies that its Euclidean norm is at most one. Apply the summing definition to $x_r=\sqrt{c_r}\,u_r$:
$$
\sum_r\|x_r\|_E^2=\sum_r c_r=n,\qquad
\sum_r|\phi(x_r)|^2=|\phi|_2^2\le1.
$$
The weak square norm is exactly one, since any contact functional attains that Euclidean norm. Hence $\pi_2(I_E)\ge\sqrt n$. Together with the upper estimate, \b[the norm is]
$$
\boxed{\pi_2(I_E)=\sqrt n.}
$$
This proves the <2-summing norm of a finite-dimensional identity> formula.
Finally retain the preceding finite-dimensional hypothesis when $E\subseteq X$. Extend each $\phi_r$ to a dual-unit functional $\widetilde\phi_r$ on $X$ using the real or complex <Hahn-Banach theorem>. Define
$$
P:X\to E,\qquad Px=\sum_r c_r\widetilde\phi_r(x)u_r.
$$
The contact identity gives $P|_E=I_E$, hence $P^2=P$ and its range is $E$. The synthesis operator $(a_r)\mapsto\sum_r\sqrt{c_r}a_ru_r$ has Euclidean operator norm one, because its product with its adjoint is the identity. Therefore
$$
\|Px\|_E\le|Px|_2\le\left(\sum_r c_r|\widetilde\phi_r(x)|^2\right)^{1/2}
\le\sqrt{\sum_r c_r}\,\|x\|_X=\sqrt n\,\|x\|_X.
$$
Thus \b[the <linear projection> satisfies $\boxed{\|P\|\le\sqrt n}$], the <Kadec-Snobar projection bound>. If $n=0$, the zero projection gives the same conclusion. No assertion about arbitrary infinite-dimensional subspaces is required.
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