Equipping $M_n$ with its usual operator space structure, $\newcommand{\ptp}{\widehat{\otimes}}$ we can form the projective tensor product of operator spaces $M_n\ptp M_n$. In particular this puts a Banach space norm on the algebraic tensor product $M_n\otimes M_n$.

Now consider the transpose map $T:M_n \to M_n$. It is a standard calculation to show that $T$ is not completely bounded, $\newcommand{\stp}{\overline{\otimes}}$ and in particular one can show that the map $\iota \otimes T : M_n \stp M_N \to M_n\stp M_n$ has norm $n$, where $\stp$ denotes the spatial tensor product (in this setting the same as the injective tensor product of operator spaces).

Question 1.What is the asymptotic behaviour (as $n\to \infty$) of $\Vert \iota \otimes T : M_n \ptp M_N \to M_n\stp M_n\Vert$?

Note that I'm asking merely about the norm as a map between two Banach spaces, not about the cb norm. If precise asymptotics are tricky, how about the following sub-question:

Question 2.In particular, does the norm of this map tend to infinity as $n\to\infty$?

This feels like something that should follow by tweaking a standard example or exercise in one of the usual books on Operator Spaces, but I couldn't succeed in converting the usual examples to get something that answers the question above.

**Remark:** the usual way to get a lower bound on
$\Vert\iota \otimes T : M_n \stp M_N \to M_n\stp M_n \Vert$ is to consider what this map does to the tensor $x = \sum_{i,j=1}^n E_{ij} \otimes E_{ji}$, the point being that $x$ has norm $1$ when viewed as an element of $M_{n^2}$ while $(\iota\otimes T)(x)$ has norm $n$ as an element of $M_{n^2}$. However, since matrix multiplication gives a complete contraction $M_n \ptp M_n \to M_n$, I think it can be shown that $x$ has norm $n$ as an element of $M_n\ptp M_n$.

**Update 2016-05-04:** I think I've now found a proof that this map is contractive for all $n$,
~~which moreover works if you replace proj tp with Haagerup tp. Previously I thought that this stronger claim (with the Haagerup tp) was false by adapting the usual argument to show the claim fails for min tp; however, this was based on a stupid miscalculation.~~
based on an interpolation argument. If the details work then I'll leave them as an answer.