Union of sets having $\binom{n - 1}{2}$ elements with each two of them having $n - 2$ common elements has cardinality of at least $\binom{n}{3}$. 
Let $S_1, S_2, \dots, S_n$ sets that have each of them $n - 1 \choose 2$ elements, with $n - 2$ common elements for each two of them. Prove that their union has at least $n \choose 3$ elements. Find an example for equality case.

Let us examine the simplest case $n = 3$. Each of them has $2 \choose 2$ $= 1$ elements and each two of them have $1$ element in common, so, in fact, the three sets are identical, so their union has exactly $3 \choose 3$ $= 1$. (Equality case confirmed)
Examining $n = 4$, each set should have $3 \choose 2$ $= 3$ elements and $2$ in common every $2$. We may construct $\{a, b, c\}$, $\{a, b, d\}$, and the last set may be $\{a, b, e\}$ or $\{a, c, d\}$, both of them satsifying the conditions.
However, I am unable to generalize the problem

One can use the following useful lemma: Let $A_1,\dots,A_N$ be $r$-element sets and $X$ be their union. If $|A_i\cap A_j|\leq k$ for all $i\neq j$, then $$|X|\geq \dfrac{r^2N}{r+(N-1)k}.$$
Proof: Let's introduce some convenient notation: $[N]:=\{1,\dots,N\}$ and for each $x\in X$ define $d(x):=\#\{j\in [N]: x\in A_j\}.$
For each $1\leq i\leq N$ consider  $$\sum \limits_{x\in A_i}d(x)=\sum \limits_{x\in A_i}\sum \limits_{j=1}^N 1_{A_j}(x)=\sum \limits_{j=1}^N |A_i\cap A_j|=|A_i|+\sum \limits_{j\neq i}|A_i\cap A_j|\leq r+(N-1)k.$$ Summing over all sets $A_i$  we have $$\sum \limits_{i=1}^N \sum\limits_{x\in A_i}d(x)\leq rN+N(N-1)k.$$ However, the LHS of the inequalit can be written as follows: $$\sum \limits_{i=1}^N \sum\limits_{x\in A_i}d(x)=\sum \limits_{i=1}^N \sum\limits_{x\in X}1_{A_i}(x)\sum \limits_{j=1}^N 1_{A_j}(x)=\sum \limits_{x\in X}\left(\sum \limits_{i=1}^N 1_{A_i}(x)\right)^2=\sum \limits_{x\in X}d(x)^2\geq $$ $$\geq \frac{1}{|X|}\left(\sum \limits_{x\in X}d(x)\right)^2=\frac{1}{|X|}\left( \sum \limits_{i=1}^{N}|A_i|\right)^2=\dfrac{N^2r^2}{|X|}.$$ Hence we obtain: $$\dfrac{N^2r^2}{|X|}\leq rN+N(N-1)k \Leftrightarrow$$ $$|X|\geq \dfrac{r^2N}{r+(N-1)k}.$$
Remark: Your problem follows easily from that lemma: just take $N=n, r=\binom{n-1}{2}$ and $k=n-2$.
