Let $Y$ denote the one-point union of three unit intervals. Since $Y$ embeds isometrically in the plane, it follows that the nth cartesian power $Y^n$ embeds isometrically — and also topologically — in $\mathbb R^{2n}$.
The metric on $Y^n$ is the cartesian power of the intrinsic metric on $Y$, so that any subset that is a product of n unit intervals has the natural metric of the unit n-cube $[0, 1]^n$.
By an "isometric embedding" of $Y^n$ into the euclidean space $\mathbb R^{k}$ I mean that $Y^n$ is isometric to some subset $X$ ⊂ $\mathbb R^{k}$ where $X$ is endowed with its intrinsic metric: D(p,q) = the infimum of the lengths of all paths from p to q within the space $X$.
If $t(n)$ is defined as the lowest dimension that $Y^n$ embeds in topologically, and $m(n)$ as the lowest dimension that $Y^n$ embeds in isometrically, then clearly we have $t(n) \le m(n) \le 2n$.
What else is known about t(n) and m(n)?
One can show that $t(Y^n)=2n$. For instance, you can use the fact that the $n$-fold join of the 3-point set does not embed in $R^{2n-2}$ due to the van Kampen obstruction, combined with Lemma 8 in
Bestvina, Mladen; Kapovich, Michael; Kleiner, Bruce, Van Kampen’s embedding obstruction for discrete groups, Invent. Math. 150, No. 2, 219-235 (2002). ZBL1041.57016.
I am quite sure it was known before and most likely proven in Wu's book "A theory of imbedding, immersion, and isotopy of polytopes in a euclidean space." (I do not have that book.) It might be also in van Kampen's original 1933 paper, but I do not have an access to it either and van Kampen's proofs were, unsurprisingly, faulty.