The authors did not say anything like "$\mathbb R^{\mathbb N}$ is not high-dimensional enough."
Rather, they said
"finite dimensional convergence [...] is too weak a result to develop genuinely high-dimensional inference methods".
The meaning here is rather the opposite: $\mathbb R^{\mathbb N}$ is "too high" dimensional for the finite-dimensional convergence to work.
Indeed, if one wants to study the behavior of $S_{n,p}:=S_n$ for large $n$ and $p$, then it is not enough to know the behavior of $S_{n,p}$ for large $n$ but only for a fixed finite set of values $p$. You already "understand that weak convergence (= convergence in distribution) in $\mathbb R^{\mathbb N}$ is equivalent to weak convergence of the finite dimensional marginals." So, the highlighted thesis follows.
On a somewhat positive note: In this comment, the OP wrote:
The fact that finite dimensional convergence implies infinite dimensional convergence sounds like a nice feature in that regard (as it simplifies things by a lot). In the paper, and in your answer, this fact sounds like I huge disadvantage though.
Of course, I said nothing of this sort. In fact, I did not talk about any advantages or disadvantages at all.
What can actually be said on this matter is the following. The convergence of the finite-dimensional distributions is of course necessary for the convergence of the distributions of the entire processes. Moreover, there are a number of results saying that, with the additional tightness condition, the convergence of the finite-dimensional distributions is also sufficient for the convergence of the distribution of the entire processes -- see e.g. Theorems 7.1 and 13.1 in Billingsley.
Furthermore, the tightness condition cannot be dropped -- cf. Example 2.7 in Billingsley's book.
Yet more, according to Prokhorov%27s_theorem, the tightness condition is necessary if the paths of the processes are in a Polish space.