I'm not aware of any natural choices for norms on $C^\infty(\mathbb{R})$ at all. (As mentioned in comments, things like the sup or $L^p$ norms are infinite for some $C^\infty(\mathbb{R})$ functions.)
There are unnatural choices, contrary to Owen Sizemore's assertion. For example, it is known that all infinite-dimensional separable Fréchet spaces (such as $C^\infty(\mathbb{R})$ in its usual topology) have Hamel dimension $2^{\aleph_0}$, hence are isomorphic as vector spaces. (I believe this is due to Mazur, but couldn't find the exact reference.) So if $(X, ||\cdot||_X)$ is your favorite infinite-dimensional separable normed space, there is a linear isomorphism $T : C^\infty(\mathbb{R}) \to X$ (of course it is in general horribly discontinuous). For $f \in C^\infty(\mathbb{R})$, set $||f|| := ||Tf||_X$. Voilà, a norm on $C^\infty(\mathbb{R})$.
Of course, under this norm $C^\infty(\mathbb{R})$ is isometrically isomorphic to $X$, so for all intents and purposes you are really working with $X$. You can even make $C^\infty(\mathbb{R})$ into a Banach space or even a Hilbert space by choosing $X$ accordingly. And the space of bounded operators on $(C^\infty(\mathbb{R}), ||\cdot||)$ is isometrically isomorphic to the space of those on $X$. Thus for any infinite-dimensional separable normed space $X$ you can realize the space of bounded operators $L(X)$ in this way, so the choice of norm on $C^\infty(\mathbb{R})$ certainly does affect the structure of $L(C^\infty(\mathbb{R}))$.
This illustrates in some sense that the norm contains almost all of the structure of a normed space; if you drop it, you are left with just the vector space structure, which determines almost nothing. "Choosing a norm" on a space is not really a sensible thing to do, since in effect it is really just choosing a normed space which could be almost totally unrelated to the original space.
Also, it should be pointed out that the space of bounded linear operators on an incomplete normed space is not a Banach space. Proof: Let $X$ be an incomplete normed space, let $\lbrace x_n\rbrace$ be a Cauchy sequence in $X$ with no limit, and let $f \in X^*$ be a nonzero bounded linear functional on $X$. Define operators $T_n$ by $T_n x = f(x) x_n$. Then $\lbrace T_n \rbrace$ is Cauchy in operator norm, but does not converge.
The correct theorem is that for normed spaces $X,Y$, the space $L(X,Y)$ of bounded linear operators from $X$ to $Y$ is a Banach space iff $Y$ is a Banach space.