First, we need to fix the notation a bit. Let $X_1,X_2,\dots$ be iid zero-mean unit-variance random variables (r.v.'s). For each natural $n$, let the $n$-tuple $(J_1,\dots,J_n):=(J_{n,1},\dots,J_{n,n})$ of r.v.'s be independent of the $X_k$'s and have the multinomial distribution with parameters $n,1/n,\dots,1/n$. Let \begin{equation*} S_n:=\frac1{\sqrt n}\,\sum_{k=1}^n J_k X_k. \end{equation*} We have to find the limit distribution of $S_n$ (as $n\to\infty$). Let us show that this limit distribution is $N(0,1/2)$. Indeed, note first here that the characteristic function (c.f.) $g_n$ of $S_n$ is given by the formula \begin{equation*} g_n(t):=Ee^{itS_n}=E\prod_{k=1}^n f(J_kt/\sqrt n) \end{equation*} for real $t$, where $f$ is the c.f. of $X_1$. Next, the joint moment generating function $M_n$ of $(J_1,\dots,J_n)$ is given by the formula \begin{equation*} M_n(t_1,\dots,t_n):=Ee^{t_1J_1+\cdots+t_nJ_n}=\Big(\frac1n\,\sum_{k=1}^n e^{t_k}\Big)^n \end{equation*} for real $t_1,\dots,t_n$. Hence, for any distinct $k$ and $l$ in $[n]:=\{1,\dots,n\}$ \begin{equation*} EJ_k^2=EJ_1^2=\frac{d^2}{dt^2}M_n(t,0,\dots,0)\Big|_{t=0}=2-1/n=2+O(1/n), \end{equation*} \begin{equation*} EJ_k^4=EJ_1^4=\frac{d^4}{dt^4}M_n(t,0,\dots,0)\Big|_{t=0}=15+O(1/n), \end{equation*} \begin{equation*} EJ_k^2 J_l^2=EJ_1^2 J_2^2=\frac{\partial^4}{\partial t^2\partial u^2}M_n(t,u,0,\dots,0)\Big|_{t=0,u=0}=4+O(1/n). \end{equation*} So, for \begin{equation*} W:=J_1^2+\cdots+J_n^2 \end{equation*} we have \begin{equation*} EW=nEJ_1^2=2n+O(1), \end{equation*} \begin{equation*} EW^2=nEJ_1^4+n(n-1)EJ_1^2 J_2^2=4n^2+O(n), \end{equation*} and hence \begin{equation*} Var\,W=O(n). \end{equation*} So, for any real $\epsilon>0$, \begin{equation*} P(|W-2n|>\epsilon n)=O(1/n)\to0, \end{equation*} so that $$\frac Wn\to2$$ in probability. Also, for the event \begin{equation} A_n:=\{\max_{k\in[n]}J_k\le n^{1/3}\} \end{equation} and its complement $A_n^c$ we have \begin{equation*} P(A_n^c)\le nP(J_1>n^{1/3})\le n\,EJ_1^4/n^{4/3}=O(1/n^{1/3})\to0 \end{equation*} and hence $P(A_n)\to1$ and $1_{A_n}\to1$ in probability. Moreover, \begin{equation*} f(s)=Ee^{isX_1}=1+is\,EX_1+(is)^2EX_1^2/(2+o(1))=1-s^2/(2+o(1))=e^{-s^2/(2+o(1))} \end{equation*} as $\mathbb R\ni s\to0$. So, for each real $t$ \begin{equation*} 1_{A_n}\prod_{k=1}^n f(J_kt/\sqrt n)=1_{A_n}\exp\Big(-\frac{t^2W}{(2+o(1))n}\Big)\to e^{-t^2} \end{equation*} in probability. On the other hand, because $|f|\le1$, \begin{equation*} \Big|1_{A_n^c}\prod_{k=1}^n f(J_kt/\sqrt n)\Big|\le1_{A_n^c}\to0 \end{equation*} in probability for each real $t$. So, by dominated convergence, \begin{equation*} g_n(t)=E\prod_{k=1}^n f(J_kt/\sqrt n) =E1_{A_n}\prod_{k=1}^n f(J_kt/\sqrt n)+E1_{A_n^c}\prod_{k=1}^n f(J_kt/\sqrt n) \to e^{-t^2} \end{equation*} for each real $t$. Thus, the distribution of $S_n$ converges to $N(0,1/2)$, as claimed.