First of all, it is easy to see that without loss of generality the dimension $n$ is $1$.
Next, of course $S:=\sum_i X_i$ can equal $\infty$ everywhere (if e.g. $X_i=1$ for all $i$), and then we will have $ES^k=\infty$ for all natural $k$.
If the $X_i$'s are independent, then $ES^k$ will be finite for all natural $k$ if and only if $\sum_i EX_i$ is finite and $\sum_i E|X_i|^k<\infty$ for all natural $k\ge2$.
This follows from Rosenthal's inequalities
$$c_1(p)(A_p+B^p)\le E\Big|\sum_i Y_i\Big|^p\le c_2(p)(A_p+B^p),$$
where $p\in[2,\infty)$, $c_1(p)$ and $c_2(p)$ are positive real constants depending only on $p$, the $Y_i$'s are independent zero-mean random variables, $A_p:=\sum_i E|Y_i|^p$, and $B:=A_2^{1/2}$.
In general, if the $X_i$'s are not independent, anything can happen. To get anything specific in the "dependent" case, you would need to specify dependence conditions on the $X_i$'s. One thing that can be said in the "dependent" case is as follows.
By Jensen's inequality, for any real $p\ge2$ and any natural $n$
$$(E|S_n|^p)^{2/p}\ge ES_n^2=(ES_n)^2+Var\, S_n,$$
where $S_n:=\sum_1^n X_i$.
So, we will have $E|S_n|^p\to\infty$ whenever $|\sum_1^n EX_i|=|ES_n|\to\infty$ (as $n\to\infty$). Also, if the $X_i$'s are nonnegatively correlated and $\sum_1^n Var\, X_i\to\infty$, then $Var S_n\ge\sum_1^n Var\, X_i\to\infty$ and hence again $E|S_n|^p\to\infty$.
