Does there always exist a continuous/smooth map from $\mathbb{R}^n$ onto a manifold $M^n$?

An obvious requirement is that $M$ must be connected. This is also sufficient.

To get a smooth surjection $f:\mathbb{R}^n\to M$ for any connected $n$-manifold $M$, cover $M$ by countably many open sets $(U_k)_{k\in\mathbb{Z}}$ which are diffeomorphic to balls in $\mathbb{R}^n$ (and such that these diffeomorphisms can be extended to smooth maps on a neighborhood of $\overline{U_n}$). Now let $f$ map $(2k,2k+1)\times\mathbb{R}^{n-1}$ diffeomorphically to $U_k$ for each $k$. On sets of the form $[2k+1,2k+2]\times\mathbb{R}^{n-1}$, we just interpolate smoothly (near the boundary, we use the fact that our diffeomorphisms extend to a neighborhood of $\overline{U_k}$, and then in between we use connectedness of $M$ to let $f$ follow some path between a point of $U_k$ and a point of $U_{k+1}$).

For continuous maps, we can do even better: we can get a continuous surjection $\mathbb{R}\to M$. The construction is the similar, except that instead of diffeomorphisms to coordinate charts we use space-filling curves.

Another construction goes through Riemannian metrics. Assume that $M$ is connected. If you put a complete riemannian metric on your manifold, the exponential map $\exp_p:T_pM\rightarrow M$ will be surjective. The idea is that any two points are connected by a geodesic. You can view the world standing in one point!