Hi Jim, maybe I can sketch an answer, assuming that I'm not doing something stupid.
The question is related to another one on this toy, namely:

http://mathoverflow.net/questions/11025/why-does-the-internal-singular-simplicial-space-realize-to-the-same-thing-as-the

By adjunction, your multisimplicial set has as its set of $(n_1,\cdots,n_k)$-simplices 
the set of continuous maps from $\Delta^{n_{1}}$ to the mapping space 
$Map(\Delta^{n_{2}} \times \cdots \times \Delta^{n_{k}}, X)$.
As $(n_2,\cdots,n_k)$ vary, these spaces form a $(k-1)$-simplicial
space, which I'll denote by $Y$.   All of the terms of $Y$ are homotopy equivalent to $X$ 
since the simplices are contractible, and it follows that the geometric realization of $Y$
is homotopy equivalent to $X$.  (See the question cited above).  Your original 
$k$-simplicial set is obtained by applying the total singular complex functor 
$S$ to each space of the $(k-1)$-simplicial space $Y$.  Realizing first in the
first simplicial coordinate we are seeing the $(k-1)$-simplicial space $|SY|$,
and its realization is homeomorphic to the realization you first asked about.  Clearly $|SY|$ is equivalent to $Y$, hence its realization is equivalent to $X$.