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 $n,1/n,\dots,1/n$. 
For each $k\in[n]:=\{1,\dots,n\}$, the value of $J_k=J_{n,k}$ is the number of times the value $X_k$ was selected into the "re-sample" from the "sample" $X_1,\dots,X_n$.  
Let 
\begin{equation*}
	S_n:=\frac1{\sqrt n}\,\sum_{k=1}^n J_k X_k,  
\end{equation*}
so that $S_n$ equals $\sqrt n$ times what you denoted by $y$. 
We have to find the limit distribution of $S_n$ (as $n\to\infty$). Let us show that this limit distribution is $N(0,2)$. 

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}=EE(e^{itS_n}|J_1,\dots,J_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 (mgf) $M_n$ 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 $t_1,\dots,t_n$. This follows because (i) the random vector $(J_1,\dots,J_n)$ is the sum of the $n$ iid random vectors $(I_{1,1},\dots,I_{1,n}),\dots,(I_{n,1},\dots,I_{n,n})$, where $I_{j,k}$ is the indicator that the value selected from the "sample" $X_1,\dots,X_n$ at the $j$th step was that of $X_k$ and (ii) the joint mgf of the random vector $(I_{1,1},\dots,I_{1,n})$ is $(t_1,\dots,t_n)\mapsto\frac1n\,\sum_{k=1}^n e^{t_k}$. 

Hence, for any distinct $k$ and $l$ in $[n]$
\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 $N(0,2)$, as claimed. 

---

That the asymptotic variance of $S_n$ is $2$ (rather than $1$, as might have been expected) stems from the fact that $EJ_k^2=2-1/n\to2$. To have another look at this phenomenon, we can let $\vec J:=(J_1,\dots,J_n)$ and write 
\begin{equation}
	Var\,S_n=E\,Var(S_n|\vec J)+Var\,E(S_n|\vec J)
	=E\frac1n\sum_1^n J_k^2+Var\,0=E J_1^2=2-1/n\to2.  
\end{equation}