Does there exist an $X\subseteq\Bbb R^3$ such that $\pi_1(X)$ is finite but nontrivial? (note that $X$ doesn't have to be a manifold or anything nice).
More generally let $f:\Bbb N\to\Bbb N$ be the function that maps $n$ to the smallest $m$ such that there is a subspace of $\Bbb R^m$ whose $\pi_n$ is finite but nontrivial, is $f(n)=n+3$? If not what can be said about this function?
I do not have an answer to your first question (and I also do not find it very natural since you are considering arbitrary subsets of $R^3$, see my comment above). I will answer the second question.
No, $f(n)\ne n+3$. For instance, $f(4)=3$ since $\pi_4(S^2)\ne 0$ and every subset of $R^2$ has trivial higher homotopy groups. For the same reason, $f(n)=3$, $n\ge 4$. Indeed, $\pi_i(S^2)$ is finite for every $i\ge 4$, and is nonzero, see
S. Ivanov, R. Mikhailov, J. Wu, "On nontriviality of homotopy groups of spheres", 2015.
Read more on homotopy groups of spheres here.
"The difference between general topology and algebraic topology is that the former asks simple questions about complicated spaces while the latter asks difficult questions about simple spaces."
Edit. Just for completeness:
For every open connected subset $U\subset R^3$, $\pi_1(U)$ is torsion-free, see my answer here.
Consequently, for every compact connected subset $K\subset R^3$ its Chech fundamental group $\check{\pi}_1(K)$ is also torsion-free. This is because $$ \check{\pi}_1(K)\cong colim_{U} \pi_1(U) $$ where the inverse limit is taken over all open neighborhoods of $K$ in $R^3$.
Therefore, for every connected CW complex $C$ (more generally, an ENR) which admits a (topological) embedding $C\to R^3$, $\pi_1(C)$ is torsion-free.