YES, trivially. Even $E(\|X_0-X_1\|^2)$ is already bounded by 4x the variance. (or even 2x ?)
Am I missing something? How isFor $n$ related$n < d$, this is optimal up to a constant factor. Take the uniform distribution on the $d$? unit vectors. (All (non-centered) moments are 1.) Then the distance is 1 if $X_0$ is distinct from $X_1,\dots,X_n$ and 0 otherwise. The expected distance is thus the probability of the first event: $(1-1/d)^n$. For $n < d$, this is between $1/e$ and $1$.
If the distribution is smooth, then $n\ge d$ makes no sense, the $n$ vectors will span the whole space with probability 1, and the expected distance is 0.
(I had also considered the uniform distribution on the sphere, but after discovering the above example, I was too lazy to calculate the expectation there, although it's just a univariate integral.