Let the elements we remove be denoted by $s^j=(s_1^j,s_2^j,\ldots,s_k^j)$ for $j=1,2,\ldots,n$. We are interested in the set $$(S_1\times S_2\times\ldots\times S_k)\setminus\{s^1,s^2,\ldots,s^n\}.$$
Claim. A (possibly very crude) upper bound is given by $\frac{n^k-1}{n-1}$.
Proof. We will prove this by induction on $k$.
For $k=1$, there is no problem. The set $S_1\setminus\{s^1,s^2,\ldots,s^n\}$ is a Cartesian product with one factor, so we are done: the upper bound is $1$ in this case.
For $k\to k+1$, note that we can rewrite the set as follows: $$\begin{align}&(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}=\\&=(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\bigcup_{x\in S_{k+1}}(S_1\times S_2\times\ldots\times S_k\times\{x\})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\left(\bigcup_{j=1}^n(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}\right)\cup(S_1\times S_2\times\ldots\times (S_{k+1}\setminus\{s_{k+1}^1,\ldots,s_{k+1}^n\})),\end{align}$$
which is (if we omit the sets that appear more than once in the first union) a disjoint union of at most $n$ sets of the form $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\},$$ and a single Cartesian product of $k+1$ sets. By the induction hypothesis, each set $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}$$ can be written as a disjoint union of at most $\frac{n^k-1}{n-1}$ Cartesian products. Therefore our set can be written as a disjoint union of at most $n\frac{n^k-1}{n-1}+1 = \frac{n^{k+1}-1}{n-1}$ Cartesian products (with $k+1$ factors each, of course). $\square$
Note that finding better bounds might require more sophisticated combinatorial methods.
Best Answer
In Wikipedia (https://en.wikipedia.org/wiki/Cartesian_product), I found something, which might be what you are looking for: $\prod_{n=1}^k \Bbb{R} = \Bbb{R}\times \Bbb{R} \times\cdots\times \Bbb{R} = \Bbb{R}^k$. So maybe something like this one is also valid: $$\prod_{\scriptstyle i = 1\atop\scriptstyle i \ne k}^nS_i$$
where $S_i$ is the $i^\text{th}$ set of the list you mentioned.