$\vee$ denotes the wedge product of the two manifolds. Let's say $p$ is the point connecting the two spheres. I would say an open neighborhood of this point is homeomorphic to an open subset of $\mathbb{R}^{n+2}$ an so $S^n \vee S^n$ could not be an $n$-dimensional manifold, but I am not entirely sure about it.
If $S^n \vee S^n$ is not a smooth manifold is it anyway possible to compute its de-Rham-cohomology?
Let $S^n_1$ and $S^n_2$ be disjoint copies of $S^n$ with basepoints $x_1$ and $x_2$ respectively, let $q\colon S^n_1 \sqcup S^n_2 \to S^n \vee S^n$ be the quotient map, and suppose $x_0 = \{x_1, x_2\}\in S^n \vee S^n$ is the wedge point. I claim there is no euclidean open set containing $x_0$, and will sketch a proof.
The definition of the quotient topology on $S^n \vee S^n$ is that $U$ is open iff $q^{-1}(U)$ is open, so if $x_0\in U$ then $q^{-1}(U)$ must be the disjoint union of non-empty open sets $U_1\subset S^n_1, U_2\subset S^n_2$ containing $x_1$ and $x_2$ respectively. In particular $U = U_1 \vee U_2$.
Now you have to argue that an open set in $S^n \vee S^n$ of the form $U_1 \vee U_2$ cannot be homeomorphic to an open $n$-ball when $n>0$ (the wedge of $0$-spheres is still a $0$-manifold, as John Palmieri points out in their comment to your question). Hint: consider the number of connected components of $U\setminus x_0$.