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 $X$.

Edit: Following up with Ricardo, the Reedy condition on Y should hold modulo the sort of elementary point-set topology that I won't look at in public (never was any good at it).  Let's look at a degeneracy operator $s_i\colon Map(\Delta^n,X)\to Map(\Delta^{n+1},X)$ for simplicity.  The multisimplicial case is no different. It is induced by a collapse map
$\sigma_i\colon \Delta^{n+1} \to \Delta^n$ which is right inverse to $\delta_i$. It follows that the image of $s_i$ is a deformation retract of $Map(\Delta^{n+1},X)$.  If there is a
continuous map $u\colon Map(\Delta^{n+1},X)\to I$ such that $u^{-1}(0) = Im (s_i)$, then
the inclusion of $Im(s_i)$ is a cofibration by the standard NDR pair criterion.  Etc.  The point is that Reedy cofibrancy is no big deal in the present context.