First, let's formulate the question properly:
Given a topological space $X$, define be $$ d(X):=\sup \{ n: X~ \hbox{is homeomorphic to} ~Y\times {\mathbb R}^n\}. $$
Lemma. The following quantities are equal when $X$ is a manifold:
(1) $d(X)$
(2) P(X):=$\max\{n: \exists ~ \hbox{a principal}~ {\mathbb R}^n\hbox{-bundle with the total space}~X\}$
(3) p(X):=$\max\{n: \exists ~ \hbox{a proper}~ {\mathbb R}^n\hbox{-action on }~X\}$, where ${\mathbb R}^n$ is equipped with the standard topology.
Proof. A principal ${\mathbb R^n}$-bundle with the total space $X$ is the same thing as a proper ${\mathbb R^n}$-action on $X$. At the same time, each principal ${\mathbb R^n}$-bundle with the total space $X$ is trivial. qed.
Remark. One can avoid using this lemma in order to justify the example below, just I find the definition $d(X)$ cleaner.
Thus, working in the topological category, you are asking if there are manifolds $X, Y$ such that $d(X\times Y)> d(X)+ d(Y)$.
Now, here is an example: Let $X$ be the Whitehead manifold (or any other contractible 3-manifold not homeomorphic to ${\mathbb R}^3$). Then $d(X)=0$. (This is not entirely trivial, but for the purpose of a counter-example we just need to know that $p(X)<3$ which is obvious since $X$ is not homeomorphic to ${\mathbb R}^3$.) On the other hand, $X\times {\mathbb R}$ is homeomorphic to ${\mathbb R}^4$ (see for instance here), hence, $d(X\times {\mathbb R})=4$.
The same example works in the smooth category.