Embedding from $L^\infty$ space to $L^2$ space

I have a set $$X$$ of $$n$$ points in a $$poly(n)$$-dimensional $$L^\infty$$ space. Does there exist a way to map the points into $$poly(n)$$-dimensional $$L^2$$ space so that the distances between points in $$X$$ are approximately preserved?

Formally, let $$S_1$$ be the original $$L^\infty$$ space with $$dim(S_1) = O(n^k)$$ ($$k$$ is known and fixed), let $$X$$ be a subset of $$S_1$$ with $$|X|=n$$, and let $$\epsilon \in (0,1)$$ be some fixed number. Does there exist an $$L^2$$ space $$S_2$$ with $$dim(S_2) = poly(n)$$ and $$\phi\colon S_1 \to S_2$$ such that $$\|\phi(x) - \phi(y)\|_2 = (1 \pm \epsilon) \|x-y\|_\infty$$ for all $$x,y \in X$$?

• Every metric space on $$n$$ points embeds isometrically into $$(n-1)$$-dimensional $$L^\infty$$.
• Embedding the metric of an $$n$$-vertex expander into $$L_2$$ requires distortion $$\Omega(\log n)$$.