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:

Why does the internal singular simplicial space realize to the same thing as the discrete singular simplicial set?

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.

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:

Why does the internal singular simplicial space realize to the same thing as the discrete singular simplicial set?

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.

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:

Why does the internal singular simplicial space realize to the same thing as the discrete singular simplicial set?

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 $X$.

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:

Why does the internal singular simplicial space realize to the same thing as the discrete singular simplicial set?

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.