I am not sure that this is what you want, but it's too long for a comment, so I post it as an answer.

I don't see serious problems. If the conditions in terms of the theory of topological vector spaces will satisfy you, then you can consider the following. Your norm $\|\cdot\|$ must

be continuous on ${\mathcal S}({\mathbb R})$ (in other words, the
unit ball $B$ of this norm must be a neighbourhood of zero in
${\mathcal S}({\mathbb R})$), and

generate a topology on ${\mathcal S}({\mathbb R})$ which is
stronger than the weak topology generated on ${\mathcal S}({\mathbb
R})$ by the duality $$ \langle f,g\rangle=\int_{\mathbb R}f(t)\cdot
g(t)\, d t,\quad f,g\in{\mathcal S}({\mathbb R}) $$ (in other words,
$\forall g\in{\mathcal S}({\mathbb R})\ \sup_{f\in{\mathcal
S}({\mathbb R}):\ \|f\|\le 1}|\langle f,g\rangle|<\infty$).

If you denote by ${\mathcal S}({\mathbb R})_{\|\cdot\|}$ the space ${\mathcal S}({\mathbb R})$ endowed with the topology generated by such a norm, then you obtain a chain of continuous mappings
$$
{\mathcal S}({\mathbb R})\subseteq{\mathcal S}({\mathbb R})_{\|\cdot\|}\subseteq {\mathcal S}'({\mathbb R}),
$$
and the completion turns it into the chain
$$
{\mathcal S}({\mathbb R})\subseteq X\to {\mathcal S}'({\mathbb R}).
$$
(since ${\mathcal S}({\mathbb R})_{\|\cdot\|}$ is normed, its completion consists of Cauchy sequences; on the other hand, by the Banach-Steinhaus theorem, ${\mathcal S}'({\mathbb R})$ is sequentially complete with respect to the ${\mathcal S}({\mathbb R})$-weak topology, so the images of these sequences have limits in ${\mathcal S}'({\mathbb R})$).

The only detail which is not clear here is if the last mapping (the arrow $\to$) is injective. In general case completion does not preserve injectivity, so you should verify this in case that this is important for you.

You can also consider the strong topology on ${\mathcal S}'({\mathbb R})$ (i.e. the topology of uniform convergence on bounded or, what is the same here, on totally bounded sets in ${\mathcal S}({\mathbb R})$), where the same reasoning works also.

thinkshould be relatively easy is that of quadratic translation-invariant norms, of the form $||u||^2=\int|\hat u(\xi)|^2\rho(\xi)\ d\xi$ $\endgroup$ – Jean Duchon Jan 27 '18 at 17:58andif there is a natural, continuous $X\subset X'$, then the dual/adjoint of the inclusion $\mathcal S\subset X$ is $X'\subset \mathcal S'$, and splicing all these together gives the picture you want. $\endgroup$ – paul garrett Jan 27 '18 at 22:53necessarycondition for the continuity of $j:X\to \mathcal S'$, namely that it factorizes over a "step" $X_n=\{u\in \mathcal S': |u(f)|\le c \sup\{(1+|x|^2)^n |f^{(\alpha)}(x)|: x\in\mathbb R^N,|\alpha|\le N\}$ for some $c\ge 0$ and all $f\in\mathcal S\}$. $\endgroup$ – Jochen Wengenroth Jan 29 '18 at 11:386more comments