This question seems to have been edited recently. The answer to the (new) question is: what you have defined is the "Banach Space Injective Tensor Product" of $X$ and $Y$. There is some information here: http://en.wikipedia.org/wiki/Topological_tensor_product
By the first part of your question, we're not actually interested in the Hilbert space representation. So we only care about what the C$^*$-algebraic minimal tensor product of $\ell^\infty(I)$ and $\ell^\infty(J)$ is. We identify these spaces with $C(\beta I)$ and $C(\beta J)$ (Stone-Cech compactifications). You'll find in standard texts that the minimal tensor product of $C(K)$ with $C(L)$ is merely $C(K\times L)$. Thus, if you follow through the calculation, you will find that for $\tau=\sum x_i\otimes y_i \in X\otimes Y$, we have
$$ \|\tau\| = \sup\Big\{ \Big| \sum_i \mu(x_i) \lambda(y_i) \Big| : \mu\in B_{X^*}, \lambda\in B_{Y^*} \Big\}. $$
This is by definition the injective tensor norm. For the basics, you can read this in Takesaki, for example (or Kadison and Ringrose I think). For more, I like Ryan's book (see the wikipedia article).
I came up with this.
Let $X$ be $I_1\times\cdots \times I_N$ tensor and $A^{(n)}$ $J_n\times I_n$ matrices. By matricization definition, $y_{j_1\cdots j_N}$ element of $J_1\times\cdots\times J_N$ tensor $Y$ maps to $y_{j_nk}$ element of matrix $Y_{(n)}$, with
$$
k=1+\sum_{\substack{l=1\\l\neq n}}^N (j_l-1)\prod_{\substack{m=1\\m\neq n}}^{l-1} J_m.
$$
We will prove the statement by showing that every element
$$
\left( X \times_1 A^{(1)} \times_2 A^{(2)}\times_3 \cdots \times_N A^{(N)}\right)_{j_1\cdots j_N}
$$
is the same as element
$$
\left[{A}^{(n)}{X}_{(n)}\left( {A}^{(N)} \otimes \cdots \otimes {A}^{(n+1)} \otimes {A}^{(n-1)} \otimes \cdots \otimes {A}^{(1)} \right)^T\right]_{j_nk},
$$
with $k$ as stated.
From $n$-mode product definition, we have
$$
\left( {X} \times_1 {A}^{(1)} \times_2 {A}^{(2)}\times_3 \cdots \times_N {A}^{(N)}\right)_{j_1\cdots j_N}=\sum_{i_1=1}^{I_1}\cdots\sum_{i_N=1}^{I_N} x_{i_1\cdots i_N} a^{(1)}_{j_1i_1}\cdots a^{(N)}_{j_Ni_N}.
$$
On the other hand, by denoting ${M}_n=\left( {A}^{(N)} \otimes \cdots \otimes {A}^{(n+1)} \otimes {A}^{(n-1)} \otimes \cdots \otimes {A}^{(1)} \right)^T$, we have
\begin{align}
\left({A}^{(n)}{X}_{(n)}{M}_n\right)_{j_nk}&={A}^{(n)}[\,j_n\,,\,:\,]\left({X}_{(n)}{M}_n\right)[\,:\,,\,k\,] \nonumber \\
&=\sum_{i_n=1}^{I_n} a^{(n)}_{j_ni_n}\left({X}_{(n)}{M}_n\right)[\,i_n\,,\,k\,] \nonumber \\
&=\sum_{i_n=1}^{I_n} a^{(n)}_{j_ni_n}\sum_{i=1}^{\hat{I}_n}{X}_{(n)}[\,i_n\,,\,i\,]{M}_n[\,i\,,\,k\,], \label{eq:proofjk}
\end{align}
with $\hat{I}_n=I_1\cdots I_{n-1}I_{n+1}\cdots I_N$. Now, ${X}_{(n)}[\,i_n\,,\,i\,]=x_{i_1\cdots i_N}$, with
$$
i=1+\sum_{\substack{l=1\\l\neq n}}^N (i_l-1)\prod_{\substack{m=1\\m\neq n}}^{l-1} I_m.
$$
The same $i$ stands in ${M}_n[\,i\,,\,k\,]$, which we easily get from the definition of Kronecker product
$$
{M}_n[\,i\,,\,k\,]=\tilde{a}^{(N)}_{i_Nj_N}\cdots \tilde{a}^{(n+1)}_{i_{n+1}j_{n+1}}\tilde{a}^{(n+1)}_{i_{n-1}j_{n-1}}\cdots\tilde{a}^{(1)}_{i_1j_1},
$$
with $\tilde{a}^{(m)}_{i_mj_m}$ denoting element of ${{A}^{(m)}}^T$. Using these conclusions, we can rewrite the above as
$$
\left({A}^{(n)}{X}_{(n)}{M}_n\right)_{j_nk}=\sum_{i_1=1}^{I_1}\cdots\sum_{i_N=1}^{I_N} x_{i_1\cdots i_N} a^{(1)}_{j_1i_1}\cdots a^{(N)}_{j_Ni_N},
$$
which completes the proof.
Best Answer
The definition on pages 14-15 of the document referenced in another answer defines this product as follows:
We will use the last line of this excerpt to make a dimensional argument about the shapes of the matrices $\mathbf U^{(n)}$ in the chain $\mathcal X \times_1 \mathbf U^{(1)} \times_2 \mathbf U^{(2)} \ldots \times_{N} \mathbf U^{(N)}$. We first clarify the meaning of a few terms. A general tensor $\mathcal X \in \mathbb{R}^{I_1 \times I_2 \times \ldots \times I_N}$ has order $N$, or equivalently has $N$ dimensions. The length of the $n$th dimension of $\mathcal X$ is $I_n$. A mode-$n$ fiber is a vector obtained by fixing all indices of $\mathcal X$ except for the $n$th dimension - if $\mathcal X$ is second order, for example, then the mode-1 fibers are the columns and the mode-2 fibers are the row.
Let's now consider an arbitrary product $\mathcal C=\mathcal X \times_n \mathbf U$ to try to understand why the result has the shape it does, and also to determine the restrictions on the shape of $\mathbf U$. The result will be a tensor of the same order as $\mathcal X$ in which the mode-$n$ fibers are multiplied (meaning standard matrix multiplication) by $\mathbf U$. Thus the length of each dimension of $\mathcal C$ will be the same as the length of the corresponding dimension of $\mathcal X$, except for the $n$th dimension, which will now have length $J$, because the mode-$n$ fibers of $\mathcal C$ are the result of multiplying the mode-$n$ fibers of $\mathcal X$ (which have $I_n$ rows) by the $J \times I_n$ matrix $\mathbf U$. Therefore, in order for the symbol $\mathcal X \times_n \mathbf U$ to make sense, we need to be able to matrix-multiply the mode-$n$ fibers of $\mathcal X$ by $\mathbf U$; i.e., the number of columns of $\mathbf U$ must be the same as the length of the $n$th dimension of $\mathcal X$.
Now let's think about how it would make sense to define something like $\mathcal X \times_n \mathbf U \times_m \mathbf V$. We need the thing on the left of the $\times_m$ symbol to be a tensor, so let's decide to first resolve $\mathcal X \times_n \mathbf U$, since we know this gives us a tensor. If $\mathcal X$ is $I_1 \times \ldots \times I_N$, then we know from the preceding that $\mathbf U$ needs to have $I_n$ columns but can have any number of rows, which we'll call $J_u$. Then $\mathcal X \times_n \mathbf U$ has dimensions $I_1 \times \ldots \times I_{n-1} \times J_u \times I_{n+1} \times \ldots \times I_N$. Now let's apply $\times_m \mathbf V$ to this. For this to make sense, the number of columns of $\mathbf V$ must agree with the length of the $m$th dimension of $\mathcal X \times_n U$, which will be $I_m$ if $n \ne m$ and $J_u$ if $n=m$.
We can apply the same reasoning to compute a chain of any length of these products, so to compute $\mathcal X \times_1 \mathbf U^{(1)} \times_2 \mathbf U^{(2)} \ldots \times_{N} \mathbf U^{(N)}$ we see by the preceding arguments that $\mathbf U^{(n)}$ needs to have $I_n$ columns and can have any number of rows; i.e., $\mathbf U^{(n)} \in \mathbb R^{J_n \times I_n}$ where $J_n$ is free.