The title says it all, but to ad a bit more:
this is mentioned here: https://en.wikipedia.org/wiki/Ordered_vector_space#Examples
is it in Bourbaki ?
does it predate the definition of total ordering (maybe by exhibiting some kind of counterexample) ? For instance, did Russell, who had deep philosophical thoughts on the topic of order, know about it ?
Many thanks! [edit: the first version of the question mentioned only "$\mathbb{R}^n$ cannot be totally ordered ($n>1$)", which is incorrect and explains the comments and answer. I'd be very interested by the answer to the reformulated question.]
The "result" for which you're looking for a citation is wrong: $\mathbb{R}^n$ can be linearly ordered for each $n\in\mathbb{N}$.
To elaborate on Brian Scott's comment: even without the axiom of choice (which per k.stm's comment implies that every set whatsoever can be ordered, and indeed well-ordered), any finite Cartesian power of an orderable set is again orderable. There are various ways to do this, with the most common probably being the lexicographic order. So since $\mathbb{R}$ is orderable each $\mathbb{R}^n$ is also orderable.
Interestingly, that "finite" bit is important: it is consistent with $\mathsf{ZF}$ (= the standard axioms of set theory without choice) that $\mathbb{R}^\mathbb{R}$ is not orderable.
Perhaps you're thinking of orderings which "play well" in some sense with the topological or algebraic structure of $\mathbb{R}^n$? For example, the usual ordering $<$ on $\mathbb{R}$ is open when viewed as a subset of $\mathbb{R}\times\mathbb{R}$, but for $n>1$ there is no ordering on $\mathbb{R}^n$ which is open as a subset of $\mathbb{R}^n\times\mathbb{R}^n$ (with all topologies in this sentence being the usual ones). I don't know when this result was proved; I suspect it's folklore.