It holds if $E[|\sum X_i|] < \infty$ or $E[\sum |X_i|] < \infty$.
Consider $X_1, X_2, ...$ in $(\Omega, \mathscr F, \mathbb P) = ([0,1], \mathscr B([0,1]), \lambda)$ where
$$X_n = 2^n 1_{A_n} + -2^{n} 1_{B_n} + 01_{A_n^C \cap B_n^C}$$ where $\lambda(A_n) = \frac{1}{2^{n+1}} = \lambda(B_n)$ and $A_i \cap B_j = A_n \cap B_n = A_i \cap A_j = B_i \cap B_j = \emptyset$
We have:
$$\sum_{n=1}^{\infty} X_n < \infty \ \lambda-a.s.$$
$$\sum_{n=1}^{\infty} E[X_n] = \sum_{n=1}^{\infty} 0 = 0$$
But we cannot compute
$$E [\sum_{n=1}^{\infty} X_n ]$$
because we have
$$ E [| \sum_{n=1}^{\infty} X_n |]$$
$$= E [|\sum_{n=1}^{\infty} 2^n 1_{A_n} + -2^{n} 1_{B_n}|] $$
$$= E [\sum_{n=1}^{\infty} |2^n| 1_{A_n} + |-2^{n}| 1_{B_n}] $$
$$= E [\sum_{n=1}^{\infty} (2^n 1_{A_n} + 2^{n} 1_{B_n})] $$
Note that $(2^n 1_{A_n} + 2^{n} 1_{B_n}) \ge 0$. Hence:
$$= \sum_{n=1}^{\infty} E [(2^n 1_{A_n} + 2^{n} 1_{B_n})] $$
$$= \infty$$
I think the above is independent of whether or not the random variables are independent.
The law of the iterated logarithm states that
\begin{align}
\limsup_n\frac{S_n}{\sqrt{n}\log\log n}\to1 \text{ a.s.},\\
\liminf_n\frac{S_n}{\sqrt{n}\log\log n}\to-1 \text{ a.s.},
\end{align}
where $S_n=\sum_1^nX_k$. In words, for each $\omega$ in a set of probability 1, the sample path $S_n(\omega)/(\sqrt{n}\log\log n)$ is arbitrary close to $1$ infinitely often and to $-1$ i.o. as $n\to\infty$. Therefore, $n^\epsilon\log\log n\to\infty$ for $\epsilon\ge 0$ implies that
\begin{align}
\limsup_n (n^\epsilon\log\log n) S_n(\omega)/(\sqrt{n}\log\log n)=\limsup_nS_n(\omega)/n^{1/2-\epsilon}=\infty\\
\liminf_n (n^\epsilon\log\log n) S_n(\omega)/(\sqrt{n}\log\log n)=\liminf_n S_n(\omega)/n^{1/2-\epsilon}=-\infty.
\end{align}
So for $p\le 1/2$ the process fails to converge almost surely.
Best Answer
It can converge to all kinds of distributions. If the common distribution is uniform on $(-1,1)$ then the characteristic function is $\phi (t)=\frac {sin (t)} t$. The characteristic function of $\sum X_n$ is $\prod \frac {sin (t/n)} {t/n}$ which has zeros. Since the characteristic function of an an infinitely divisible distribution never vanishes it follows that the the sum has a distribution that is not even infinitely divisible, so certainly not normal.