There are at most $444$ subsets. Assume $|V|=1000$ and we have $N$ three-element subsets $V_1,V_2,...,V_N$ satisfies $\forall 1\leqslant i \leqslant N,V_i\subset V$,and $\forall 1\leqslant i_1<i_2<i_3<i_4<i_5\leqslant N,\left|\bigcup_{j=1}^5 V_{i_j}\right|\geqslant 12.$ We construct a graph $G$ with the vertice $v_i$ representing the set $V_i$.Let $V(G)$ be the set of the vertices,$E(G)$ be the set of the edges,$d(v_i)$ be the degree of $v_i$.For two distinct vertices $v_i,v_j$,we connect $v_i,v_j$ with $1$ edge if $|V_i\cap V_j|=1$;and connect $v_i,v_j$ with $2$ edges if $|V_i\cap V_j|=2$;and draw no edge between $v_i,v_j$ if $|V_i\cap V_j|=0$. Thus we can get: (1)Any induced graph of $G$ with order $5$ has at most $3$ edges; (2)$|E(G)|\geqslant 3N-1000$. Case1:There exist two vertices in $G$ with $2$ edges between them,say $v_1,v_2$. Case1.1:$d(v_1)+d(v_2)>4$. Since $\forall v_i\in V(G),d(v_i)\leqslant 3$,WLOG let $d(v_1)=3$ and the other neighbor of $v_1$ is $v_3$.So from (1) we know $d(v_3)=1$ and $\forall 4 \leqslant j_1<j_2 \leqslant N,v_{j_1}v_{j_2} \notin E(G)\Rightarrow |E(G)|=3\Rightarrow N\leqslant 334.$ Case1.2:$d(v_1)=d(v_2)=2$. From (1) we can get that $\forall 3\leqslant j\leqslant N,d(i_j)\leqslant 1$,Thus $|E(G)|\leqslant \left \lfloor \frac{N-2}{2}\right \rfloor+2\leqslant \frac{N+2}{2}$.Combine with (2) we can get that $N\leqslant 400.$ Case2:For any two vertices in $G$ there is at most $1$ edge between them. Case2.1:There exists a cycle in $G$. From (1) we know that the cycle must be a triangle,say $v_1v_2v_3$.Then from (1) we can get that $d(v_1)=d(v_2)=d(v_3)=2$ and $\forall 4 \leqslant j_1<j_2 \leqslant N,v_{j_1}v_{j_2} \notin E(G)$.Then by Case1.1 we can get the same bound $N\leqslant 334$. Case2.2:There are no cycles in $G$.So $G$ is a forest. Then from (1) we know that the order of a tree in $G$ is no more than $4$.Assume that the number of $k$-order tree in $G$ are $x_k~(k=1,2,3,4)$. Then:$N=x_1+2x_2+3x_3+4x_4,|E(G)|=x_2+2x_3+3x_4=N-x_1-x_2-x_3-x_4\leqslant N-\frac{1}{4}N=\frac{3}{4}N$. So combine with (2) we can get:$\frac{9}{4}N\leqslant 1000\Rightarrow N\leqslant 444.$ Combine all the cases we get that $N\leqslant 444$.Easy to verify that if we let $G$ consist of $111$ nonintersecting $4$-order tree,then $G$ satisfies (1) and (2). So there are at most $\boxed{444}$ three-element subsets.Done!

Lemma 1: If for a fixed natural number $n$ we have at least $n$ $\text{three-element}$ subsets of a set and the union of every $n$ of them has at least $2n+2$ elements, the subsets can be divided into some families such that if a family has $t$ sets, we can imply $t<n$ and also the union of the family has at least $2t+1$ elements. And any two sets from two distinct families don't have any intersection. Proof: I prove it with induction on $n$. For $n=1$ every set should have at least $4$ elements and we're done. If $n\geqslant2$, assume that there exist at least $n$ subsets. While there exist $n-1$ subsets such that the union of them has at most $2n-1$ elements, call them a family. Considering the sets of that family and another subset, the problem's assumptions imply that the union of that family should have exactly $2n-1$ elements and also the sets in the family don't have any intersection with other subsets. Do this until the subsets are divided into families with $n-1$ sets and $2n-1$ elements and some subsets like $\mathcal{A}$ that the union of every $n-1$ of them has at least $2n$ elements. Here we have 2 cases: Case 1: $\lvert\mathcal{A}\rvert\geqslant n-1$ Let the union of all the families I constructed above $\mathcal{F}_{n-1}$ and by induction hypothesis, we can write $\mathcal{A}=\bigcup_{t=1}^{n-2}\mathcal{F}_t$ such that $\forall t$ $\mathcal{F}_t$ is union of all the families with $t$ sets. Hence, we are done. Case 2: $\lvert\mathcal{A}\rvert<n-1$ In this case, name $\lvert\mathcal{A}\rvert=t$. Case 2.1: $\lvert\cup\mathcal{A}\rvert\geqslant2t+1$ The families I mentioned above with $n-1$ sets and $\mathcal{A}$ together cover all the subsets. Case 2.2: $\lvert\cup\mathcal{A}\rvert\leqslant2t$ We know that there exist at least $n$ subsets but $\lvert\mathcal{A}\rvert<n$ so there exists at least $1$ family with $n-1$ sets and exactly $2n-1$ elements. If the whole number of subsets is $s$ and the number of distinct elements used in the subsets is $l$, by recent sentences we can conclude $l=\frac{2n-1}{n-1}(s-t)+\lvert\cup\mathcal{A}\rvert$. But now we divide $\mathcal{F}_{n-1}$ in this way: Considering the sets of $\mathcal{A}$ and every other $n-t$ sets in $\mathcal{F}_{n-1}$, the problem's assumptions imply that the union of every $n-t$ sets in $\mathcal{F}_{n-1}$ has at least $2n-2t$ elements so by the induction hypothesis, $\mathcal{F}_{n-1}$ can be divided into families with at most $n-2$ sets. So $l\geqslant\frac{2n-3}{n-2}(s-t)+\lvert\cup\mathcal{A}\rvert\implies s-t=0$. Contradiction. So this case never occurs. So by the lemma since $n=5$ if the set of subsets be $\mathcal{S}$ we can write $\mathcal{S}=\bigcup_{t=1}^{4}\mathcal{F}_t$ such that $\forall t$ $\mathcal{F}_t$ is union of all the families with $t$ sets. Hence we have: $$\lvert\mathcal{S}\rvert=\left\lvert\bigcup_{t=1}^{4}\mathcal{F}_t\right\rvert=\sum_{t=1}^{4}\lvert\mathcal{F}_t\rvert\leqslant\sum_{t=1}^{4}\frac{t}{2t+1}\lvert\cup\mathcal{F}_t\rvert\leqslant\frac{4}{9}\sum_{t=1}^{4}\lvert\cup\mathcal{F}_t\rvert\leqslant\frac{4}{9}\times1000\implies\boxed{\lvert\mathcal{S}\rvert\leqslant444}$$ Now I present an example for $444$. $\forall 0\leqslant k<111$ consider the $k$-th family having the following sets: $$\{9k+1,9k+2,9k+3\},\{9k+1,9k+4,9k+5\},\{9k+1,9k+6,9k+7\},\{9k+1,9k+8,9k+9\}$$$\forall 1\leqslant t\leqslant4$ the union of any $t$ sets of family has exactly $2t+1$ elements (the number $9k+1$ and the $2t$ distinct other numbers). So, the number of elements of the union of every $5$ subsets is exactly $10$ plus the number of distinct families they belong to, which is at least $12$ (because every family has at most $4$ sets) and we're done$\blacksquare$

Mehrshad wrote: $\forall 0\leqslant k<111$ consider the $k$-th family having the following sets: $$\{9k+1,9k+2,9k+3\},\{9k+1,9k+4,9k+5\},\{9k+1,9k+6,9k+7\},\{9k+1,9k+8,9k+9\}$$$\forall 1\leqslant t\leqslant4$ the union of any $t$ sets of family has exactly $2t+1$ elements (the number $9k+1$ and the $2t$ distinct other numbers). So, the number of elements of the union of every $5$ subsets is exactly $10$ plus the number of distinct families they belong to, which is at least $12$ (because every family has at most $4$ sets) Now all $4$ sets I mentioned above as sets of a family, are pairwise connected in your graph. So those $ 4$ vertices: Form a clique with a size of $4$ Any induced graph of $G$ having those $ 4$ vertices has at least $6$ edges. It's possible to make a cycle with a size of $4$ with those $4$ vertices.