"Identity tensor transpose" as a map $M_n \hat{\otimes} M_n \to M_n \overline{\otimes} M_n$ 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.
 A: Well I guess I should write something quickly, even if it doesn't have all the details, otherwise I'll keep putting it off. And maybe someone will spot a mistake...
Fix Hilbert spaces $V$ and $W$, which we think of as having column OSS. Equip $B(V)$ and $B(W)$ with their usual OS structures. Then the linear map 
$$ \iota\otimes \top : B(V) \otimes B(W) \to B(V \otimes_2 W)$$
extends to a contractive linear map $B(V) \ptp B(W) \to B(V\otimes_2 W)$.
The proof goes in stages.
Step 1. If $x = \sum_i a_i \otimes b_i \in B(V)\otimes B(W)$, then 
$$ \Vert \sum\nolimits_i a_i \otimes b_i^\top \Vert_{B(V\otimes_2 W)}
 = \sup \left\{ \Vert \sum\nolimits_i a_i c b_i \Vert_{HS(W,V)}
\;\colon\; c \in HS(W,V), \Vert c\Vert_{HS(W,V)} \leq 1 \right\} $$
(This is not hard to hack out by hand, but with hindsight can also be found in various places, for instance I think it is in Pisier's book somewhere early on.)
Step 2. Note that there are have natural completely contractive maps$\newcommand{\ptp}{\widehat{\otimes}}$
$$ B(V) \ptp V \to V\quad,\quad W^* \ptp B(W) \to W^* $$
where the first is the usual action $a\otimes v \mapsto av$ and the second is the transposed action $\psi\otimes b \mapsto \psi\circ L_b$, $L_b$ being the action $w\mapsto bw$. Therefore by general stuff on operator space tensor products, we have complete contractions$\newcommand{\itp}{\otimes_{\rm min}}$
$$ B(V) \ptp( V\ptp W^*) \ptp B(W) \to V\ptp W^* $$
$$ B(V) \ptp( V\itp W^*) \ptp B(W) \to V\itp W^* $$
where $\itp$ denotes injective tensor product of operator spaces.
Note that if we identify $V\otimes W^*$ with the space of finite rank operators $W\to V$, then the two maps above just correspond to $a\otimes c \otimes b \mapsto acb$.
Step 3. Under the natural identification of $V\otimes W^*$ with the finite-rank operators $W\to V$, we have isometric isomorphisms $V\ptp W^*\cong S_1(W,V)$ and $V\itp W^* \cong S_\infty(W,V)$, where $S_1$ denotes trace-class operators and $S_\infty$ the compact operators.
Step 4. Let $x$ be as in Step 1 and WLOG assume its norm in $B(V)\ptp B(W)$ is $1$. Let $E_x$ denote the elementary operator on $B(W,V)$ defined by $c\mapsto \sum_i a_icb_i$. We wish to show that the norm of $E_x$ as a map $HS(W,V)\to HS(W,V)$ is $\leq 1$. But now by Steps 2 and 3, we know that $E_x$ is (completely) contractive from $S_1(W,V)$ to $S_1(W,V)$, and (completely) contractive from $S_\infty(W,V)$ to $S_\infty(W,V)$. By classical complex interpolation results, $E_x$ is therefore contractive on all the intermediate Schatten classes, in particular on the Hilbert-Schmidt operators, and we are done.
Remark. Of course we can interpolate in the category of operator spaces. The argument above, if correct, seems to actually show that $E_x$ is completely contractive on $HS(W,V)$ when we equip this space with the "self-dual" OSS. If I denote this operator space by OH temporarily, then we can rephrase this as: $\iota\otimes \top$ is contractive from $B(V)\ptp B(W)$ to $CB(OH)$. It seems plausible that we actually get a complete contraction, but I haven't yet done the book-keeping required to check this.
A: 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.)
EDIT:  The above argument is faulty: in particular, it's not clear to me that the 1st level norm on the operator space tensor product $M_n\widehat{\otimes}M_n$ coincides with the projective tensor product of $M_n$ with $M_n$ in the category of Banach spaces. (There's a remark to this effect for general $X\widehat{\otimes}Y$ in the Blecher-Paulsen paper. However I'm not sure whether or not its true in the particular case of $X=Y=M_n$ with the usual operator space structure; I'll leave the above argument in place for now in case someone else knows how to fix it.)
