Here is a simple proof that the property holds only for Euclidean norms, at least if the norm in question is $C^1$ smooth and strictly convex. Surely it was known way before Gromov was born.
Let $S$ denote the unit sphere of the norm. First observe that, if $v\in S$ and $w$ are such that $\|w+tv\|\ge \|tv\|$ for all $t\in\mathbb R$, than $w$ is parallel to the tangent plane to $S$ at $v$.
So we have a map $F$ from the set of 2-dimensional linear subspaces to the set of 1-dimensional subspaces such that, for every linear plane $P$, the line $F(P)$ is parallel to the tangent planes at all points of $S\cap P$. Smoothness and strict convexity imply that this map is continuous and injective, hence it is bijective.
If a collection of planes contain a common line (or, equivalently, a common unit vector $v$), then their $F$-images belong to one plane (namely the one parallel to the tangent plane to $S$ at $v$). Now consider the set of planes and the set of lines as models of the real projective plane (they are the projectivizations of $(\mathbb R^3)^*$ and $\mathbb R^3$, resp.) Now we have a bijection from the projective plane to itself that sends lines to lines, hence a projective map.
Thus there is a linear map
$L:(\mathbb R^3)^*\to\mathbb R^3$ whose projectivization is $F$. This means that, if a plane $P$ is the kernel of some element $p\in(\mathbb R^3)^*$, then $L(p)$ is a vector from the line $F(P)$. This map $L$ defines a non-degenerate bilinear form on $\mathbb R^3$ in a standard way, and $F(P)$ is just the orthogonal complement to $P$ w.r.t. this form. Since $F(P)$ obviously never belongs to $P$, the bilinear form is sign-definite (assume positive), so it is a Euclidean structure. In this Euclidean structure, the tangent planes to $S$ are orthogonal to their corresponding radial vectors. It follows that $S$ is a sphere of this Euclidean structure.