Note that there is a distance preserving map between Mahattan norm and sup norm : If $A : (\mathbb{R}^2,\|\ \|_1) \rightarrow (\mathbb{R}^2,\|\ \|_\infty)$ is a linear map by $$ A(e_1)=e_1+e_2 , \ A(e_2)= -e_1+e_2 $$ where $e_i$ is a canonical basis, then $A$ is distance preserving by a direct computation.
But prove that there is no bijective distance preserving map $f$ with $f(o)=o$ between $(\mathbb{R}^3,\|\ \|_1) \rightarrow (\mathbb{R}^3,\|\ \|_\infty)$ :
If $\Sigma_1$ is $\|\ \|_1$-unit sphere, and $d_1$ is intrinsic metric on $\Sigma_1$, then there is six points $x_i\in \Sigma_1$ s.t. pairwise distance is at least $2$.
Similarly, if we have $\Sigma_\infty,\ d_\infty$, then there is eight points $y_i$ s.t. pairwise distance is at least $2$. Hence I think that if we cannot find $z_i\in \Sigma_1,\ 1\leq i \leq 8$ s.t. pairwise distance is at least $2$, then we can complete the proof. How can we show ?
If there is another approach, I could admit it.
[Add] If so is $f$, then $f$ should be a linear map ?
In $(\mathbb{R}^3,\|\ \|_1)$ for $x=(x_i)$ with $\|x\|_1$, $x$ is a canonical basis iff there is only one shortest path from the origin to $x$ : For instance, $x=(x_1,x_2,0),\ x_i\neq 0$, then $c(t)=tx$ is a shortest path. And union of two paths $a(t)=t(x_1,0,0),\ b(t)=(x_1,tx_2,0)$ is also a shortest path. Similary, in $(\mathbb{R}^3,\|\ \|_\infty)$, there are exactly eight points that have unique shortest paths. Since bijective distance preserving map sends a point having unique shortest path to a point having unique shortest path, then there can not exist such a map.