The lower bound argument I gave for $\sqrt{n}$ points in a square works here, too. I have tried to simplify it. The idea is to use the union bound: The probability that a random path with $m=\lfloor \sqrt[3]{n} \rfloor$ steps has length less than some constant is small. It is so small that the expected number of ways to choose a short path is less than $1$, so the probability is less than $1$, in fact with probability going to $1$ there is no such short path using $m$ vertices out of $n$.

It helps to suppose the points are chosen from a fine lattice $\{1/\ell,2/\ell,...\}^3$ with $\ell \gg m$, and to estimate the probability that the sum of the $L^1$ distances along the path is small rather than the $L^2$ distance. Pushing the points to be on a fine lattice does not change the length much, less than $2\sqrt{3}m/\ell$.

The count of sequences of $m$ nonnegative steps of total $L^1$ length up to $d$ is the number of ways of distributing $d$ objects among $3m+1$ categories, $d+3m \choose 3m$. There are at most $2^{3m}$ choices of signs. So, the probability that a random path with $m$ steps has $L^1$ length at most $c \ell$ is at most $2^{3m}{c \ell +3m \choose 3m} /\ell^{3m} \le \frac{2^{3m}(c \ell + 3m)^{3m}}{(3m)!\ell^{3m}} = \frac{(2c+6m/\ell)^{3m}}{(3m)!}$. If $c \ell \gt 3m$ we can estimate this as less than $\frac{(4c)^{3m}}{(3m)!}$. (Actually, we don't need to accept this factor of $2$.) Using $x! \gt (x/3)^x$, this is less than $\left(\frac{4c}{m}\right)^{3m}$.

The number of ways to choose an $m$ step path from $n$ points is at most $n\times(n-1)\times(n-2)...\times(n-m) \le n^{m+1} \approx m^{3m+3}.$

The expected number of paths with $m$ steps of $L^1$ length less than $c$ is at most $m^{3m+3}\left(\frac{4c}{m} \right)^{3m} =m^3 (4c)^{3m}$. So, if we choose $c \lt 1/4$ then as $m,n \to \infty$, the probability that there is a path with $L^1$ length smaller than $c$ goes to $0$.

The Euclidean length is up to $\sqrt{3}$ times smaller than the $L^1$ length, so the probability that there is a path through $\sqrt[3]{n}$ points with Euclidean length less than $1/(4\sqrt{3})$ goes to $0$ as $n \to \infty$.

The constant can be improved easily by a factor of $2$, and with a little work one can estimate the probability that the $L^2$ distance is small directly instead of the $L^1$ distance, and this should improve the constant significantly, too. The convenient estimate $x! \gt (x/3)^x$ can be improved to a lower bound of roughly $(x/e)^x$ from Stirling's formula which improves the constant a bit more.