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 changed 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 $\chi(\mathbb{C} \mathbb{P}^{2} ) = 3$. So for any $4$-manifold $N$ connect summing with $\mathbb{R}\mathbb{P}^{2} \times \mathbb{R}\mathbb{P}^{2}$ subtracts 1 from $\chi(N)$, connect summing with $\mathbb{C} \mathbb{P}^{2} $ adds one to $\chi(N)$, 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.
Edit 1: After reading the question more carefully, I realise this is not a complete solution. The embedding we obtain does not have dense image after doing the connect sums. (Although see the comment of Misha below).
Edit 2. Note that the above solution holds in even dimensions atleast $4$. Let me construct a counterexample in the dimension $2$ case.
Let $S$ be a genus $3$ surface. I will show there is not embedding of $S \setminus \{p\}$ to $\mathbb{T}^{2}$ (for some $p \in S$).
Write $S$ as $\mathbb{T}^{2} \# \mathbb{T}^{2} \#\mathbb{T}^{2}$, and let $C \subset S$ be the cuve along which the first connect sum is made. Then if we contract $C$ to a point in $S$, we obtain the wedge sum of a genus $2$ surface and a torus. We set $p$ so that in the image of this contraction it will lie in the torus.
Suppose there is an ebbeding $I: S \setminus \{p\} \hookrightarrow \mathbb{T}^{2} $, then If we contract $I(C)$ we obtain the wedge sum of a genus $2$ surface with some other surface. One can see there is not simple close curve on the torus with this property. (Indeed for any simple closed curve on the torus the space obtained by contracting this curve will be a "pinched torus" or $S^{2} \vee \mathbb{T}^{2}$)
(However, I don't know how to show a puctured genus 2 surface doesn't embed in a torus).