I have computed the number of monomials of the polynomial:
$$\prod_{0 \le k_1 \lt \ldots \lt k_{n-t} \le n-1} \left( \sum_{j=1}^{n-1} x^{k_j} \right)$$
with $t = 1$ for $n=2,3,4,5,6$ and found it equal to $1,5,11,19,29$ respectively.
Each factor is a combination of $n-t$ monomials chosen among $1,x,x^2,\ldots,x^{n-1}$. The product is over all possible combinations.
For example, with $n=3$ we have:
$$P(x) = (1+x)(1+x^2)(x+x^2) = x^5 + 2 x^4 + 2 x^3 + 2 x^2 + x$$
and therefore there are $5$ monomials.
Looking at OEIS, one could for example conjecture that the number of monomials is:
$$n-2+(n-1)^2 = n^2-n-1$$
Is the conjecture true? In general, is it possible to find a formula as a function of $n$ for any given $t$ such that $n \gt t \ge 1$?
We simply need to find the highest degree that appears. To find this, note that when $t=1$, each sum of monomials will only remove one $x^i$ term, where $0\leq i \leq n-1$. As there are ${n \choose n-1}=n$ such terms, $n-1$ with highest degree $n-1$ and 1 with highest degree $n-2$ (which is precisely the term $\sum_{i=0}^{n-2} x^i$), the term with the highest degree appearing is $x^{(n-1)^2+n-2}$.