No, it is not; here is a slightly simpler argument than the one indicated in my comment. As in the answer to the previous question, let $X=\Delta_3/\partial\Delta_3$. It suffices to show that the cardinality of $A_n=\operatorname{Hom}(\Delta_1\times \Delta_n, X)$ grows faster than any polynomial as a function of $n$. As in that answer, an element of $A_n$ can be identified with a sequence $(x_0,\dots,x_n)$ of elements of $X_{n+1}$ such that $d_{i+1}x_i=d_{i+1}x_{i+1}$ for $0\leq i<n$.

Let $B_n\subset A_n$ be the set of such sequences for which $d_{i+1}x_i=[\partial\Delta_3]$ is the basepoint of $X_n$ for all $0\leq i<n$ and $x_0=x_n=[\partial\Delta_3]$ is the basepoint of $X_{n+1}$. If $(x_0,\dots,x_n)\in B_n$, then for each $0<i<n$, $x_i$ must be either $[\partial\Delta_3]$ or the (image in $X$ of) the unique $(n+1)$-simplex $\sigma_i$ in $\Delta_3$ such that $\sigma_i$ is not in $\partial\Delta_3$ but $d_i\sigma_i$ and $d_{i+1}\sigma_i$ are in $\partial\Delta_3$. Explicitly, this simplex is the map $\sigma_i:[n+1]\to[3]$ such that $\sigma_i(j)=0$ for $j<i$, $\sigma_i(i)=1$, $\sigma_i(i+1)=2$, and $\sigma_i(j)=3$ for $j>i+1$.

So to define an element of $B_n$, we have exactly two choices for each $x_i$ for $0<i<n$, and every possible sequence of such choices does define an element of $B_n$. Thus $B_n$ has exactly $2^{n-1}$ elements. In particular, $|A_n|\geq 2^{n-1}$ grows exponentially.