About the identification of $M_n(B(\mathcal{H}))$ with $B(\mathcal{H}^{(n)})$ I already checked another question where the *-isomorphism is shown.
As said in the book I'm reading now Completely bounded maps and operator algebras by Vern Paulsen page 3:
"There is a natural way to regard an element of $M_n(B(\mathcal{H}))$ as a linear map on $\mathcal{H}^{(n)}$, by using the ordinary rules for matrix products. That is, we set:
$$ (T_{ij}) \begin{pmatrix}h_1 \\
 \vdots\\ h_n \end{pmatrix}  = \begin{pmatrix}\sum_{j=1}^nT_{1j}h_j\\
\vdots\\ \sum_{j=1}^nT_{nj}h_j \end{pmatrix}$$
fot $(T_{ij})$ in $M_n(B(\mathcal{H}))$ and $(h_1, \dots h_n)^T \in \mathcal{H}^{(n)}$..".
What we should check now is that every element of $M_n(B(\mathcal{H}))$ defines a bounded linear operator on $\mathcal{H}^{(n)}$ obtaining the following inequality:
$$\|(T_{ij})\| \le \left(\sum_{i,j=1}^n\|T_{ij}\|^2\right)^{1/2}$$
Any help on how to do that?
EDIT:
That's what I'm thinking to do right now:
take $\bar{h}=(h_1, \dots, h_n) \in \mathcal{H}^{(n)}$ and consider
$$ \|(T_{ij})(\bar{h})\|^2 = \sum_{i=1}^n \left\|\sum_{j=1}^n T_{ij}h_j\right \|^2 \le \sum_{i,j = 1}^n \|T_{ij}h_j\|^2 \le \sum_{i,j =1}^n \|T_{ij}\|^2 \|h_j\|^2 $$
how to formally conclude?
Also the fact of considering elements of $\mathcal{H}^{(n)}$ as row vectors instead of columns vectors helped me quite. There would have been any difference?
Also, showing the isomorphism, I'd need some help for surjectivity : /
 A: Your first inequality is not true, the square of the norm does not satisfy the triangle inequality. What you can do is use the triangle inequality and then Cauchy=Schwarz as follows:
$$
\left\|\sum_{j=1}^n T_{ij}h_j\right \|^2
\leq\left(\sum_{j=1}^n \|T_{ij}\|\,\|h_j\|\right )^2\leq\sum_{j=1}^n\|T_{ij}\|^2\sum_{j=1}^n\|h_j\|^2.
$$
Then
$$\sum_{i=1}^n
\left\|\sum_{j=1}^n T_{ij}h_j\right \|^2
\leq\sum_{i=1}^n\left(\sum_{j=1}^n\|T_{ij}\|^2\sum_{j=1}^n\|h_j\|^2\right)
=\left(\sum_{i,j=1}^n\|T_{ij}\|^2\right)\sum_{j=1}^n\|h_j\|^2
$$
If your vectors are rows, the multiplication by a matrix is not "row times column", and it doesn't work on the left, the matrix $A$ acting on $x$ would be $xA^T$ (depending on how you define the matrix of an operator). 
For the surjectivity, if $T\in B(H^n)$, if $\pi_j:H^n\to H$ is the projection onto the $j^{\rm th}$ coordinate, you can define $$T_{kj}h=\pi_k \circ T\begin{bmatrix} 0\\ \vdots \\ 0\\ h\\ 0\\ \vdots\\ 0\end{bmatrix},$$ where the $h$ is in the $j^{\rm th }$ coordinate. 
