First note that odd dimensions the question of Euler characteristic $0$ is automatic, $M$ will embed in the orientable double cover of $\tilde{M}$, which will have $\chi = 0$ by Poincare Duality.
In even dimension = $2n$ (we assume $n > 1$), we recall the following fact. If $M_{1},M_{2}$ are compact connected manifolds then $\chi(M_{1} \# M_{2}) = \chi(M_{1}) + \chi(M_{2}) - \chi(S^{2n}) = \chi(M_{1}) + \chi(M_{2}) - 2$.
To prove that some embedding into any manifold implies embedding in to Euler characteristic $0$ manifold, it is sufficient to show that any integer is equal to the Euler characteristic of some manifold of dimension $2n$, since we can do connect sums on the complement of the embedding $M \hookrightarrow \tilde{M}$ (It is easy to see that the embedding can be made so that this complement contains an open set) to shift the Euler characteristic of $\tilde{M}$ to the correct value.
In dimension $4$ we have $\chi(\mathbb{R}\mathbb{P}^{2} \times \mathbb{R}\mathbb{P}^{2}) =1$ and $\chi(\mathbb{C} \mathbb{P}^{2} ) = 3$. Connect summing with $\chi(\mathbb{R}\mathbb{P}^{2} \times \mathbb{R}\mathbb{P}^{2})$ subtracts 1 from $\chi$, connect summing with $\chi(\mathbb{C} \mathbb{P}^{2} )$ adds one to $\chi$, hence there is a $4$-manifold with Euler characteristic equal to any integer. In higher even dimension taking appropriate products with $\mathbb{R}\mathbb{P}^{2}$ will give the same result.