My gut feeling is that these maps should have norm 1 for all $n$, here is an attempted argument:
For $n$ fixed and $k\geq 1$ we identify the algebraic tensor product $M_k\otimes M_n$ with $M_k(M_n)$ in the usual way. If we fix an operator space structure on $M_n$, then for each $k$ the norm on $M_k(M_n)$ induces a norm on $M_k\otimes M_n$. If I'm not mistaken, it's a result of Blecher and Paulsen [Tensor Products of Operator Spaces, JFA 1991] that the map $$ M_n\widehat{\otimes}M_n\to M_n\otimes_{max} M_n $$ is contractive (as a map between Banach spaces), where $max$ is the norm on $M_n\otimes M_n\cong M_n(M_n)$ induced by the maximal operator space structure on $M_n$. On the other hand, if we give $M_n$ its maximal operator space structure, then every contractive map $T:M_n\to B(H)$ is completely contractive, so the map in question factors as $$ M_n\widehat{\otimes}M_n\to M_n\otimes_{max} M_n \to M_n\overline{\otimes}M_n $$ where the first map is the identity, and the second is $id\otimes T$. But $M_n\overline{\otimes}M_n$ is the norm on $M_n(M_n)$ induced by the minimal operator space norm, and hence $id\otimes T$ is contractive (since $T$ is completely contractive out of the $max$ norm.)