The answer to your question is known as Katonas Theorem.
We use the following notation: Let $\mathcal{N}$ be the set of positive integers, $[i,j]=\{i,i+1,\dots,j\}$ for $i,j \in \mathcal{N}, [n] = [1,n]$ and for $k<n$ let $2^{[n]}=\{A:A\subset[n]\}$ be the unrestricted subsets of $[n]$.
A system of sets $\mathcal{A}\subset 2^{[n]}$ is called k-intersecting, if $|A_1 \cap A_2| \geq k$ for all $A_1,A_2\in \mathcal{A}$.
Theorem (Katona $1964$):
Let $I(n,k)$ denote the set of all $k$-intersecting systems and $M(n,k)=max_{\mathcal{A}\in I(n,k)}|\mathcal{A}|$. The following statement is valid
\begin{equation*}
M(n,k)=
\begin{cases}\sum_{j=\frac{n+k}{2}}^{n}\binom{n}{j},&2 \mid (n+k)\\
&\\
2\sum_{j=\frac{n+k-1}{2}}^{n-1}\binom{n-1}{j},&2 \nmid (n+k)
\end{cases}
\end{equation*}
The answer above together with a really short proof can be found in the paper Katona's intersection theorem: Four proofs, from R. Ahlswede and L. H. Khachatrian.
Note: An interesting overview with answers to your and related questions can be found in Intersecting families of sets and permutations: a survey from Peter Borg ($2011$).
Added 2014-07-19: I've added below a table and examples for small values $M(n,k)$ with $n \leq 6$ to provide a first impression which $k$-intersecting families $\mathcal{A}$ have size $|\mathcal{A}| > 2^{n-k}$.
The interesting values $M(n,k) > 2^{n-k}$ $(2\leq n \leq 6$ and $1 \leq k \leq n-1)$ are written $\bf{bold}$.
$M(n,k)$:
\begin{align*}
\begin{array}{crrrrr}
n/k&1&2&3&4&5\\
\hline\\
2&2\\
3&4&2\\
4&8&\bf{5}&2\\
5&16&\bf{10}&\bf{6}&2\\
6&32&\bf{22}&\bf{12}&\bf{7}&2
\end{array}
\end{align*}
The following four examples are simply for illustration. For $$(n,k)\in\{(4,2),(5,2),(5,3),(6,2)\}$$ a family $\mathcal{A}_1$ according to the appoach of talegari with $|\mathcal{A}_1|=2^{n-k}$ elements and a family $\mathcal{A}_2$ with $|\mathcal{A}_2|=M(n,k) > 2^{n-k}$ is listed.
Example: $n=4,k=2$
$|\mathcal{A}_1|=2^{4-2}=4$, $|\mathcal{A}_2|=M(4,2)=5$
\begin{align*}
\begin{array}{rl}
\mathcal{A}_1=&\{\{1,2\},\\
&\{1,2,3\},\{1,2,4\},\\
&\{1,2,3,4\}\}\\
\end{array}
\qquad
\begin{array}{rl}
\mathcal{A}_2=&\{\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\\
&\{1,2,3,4\}\}\\
\\
\end{array}
\end{align*}
Example: $n=5,k=2$
$|\mathcal{A}_1|=2^{5-2}=8$, $|\mathcal{A}_2|=M(5,2)=10$
\begin{align*}
\begin{array}{rl}
\mathcal{A}_1=&\{\{1,2\},\\
&\{1,2,3\},\{1,2,4\},\{1,2,5\},\\
&\{1,2,3,4\},\{1,2,3,5\},\{1,2,4,5\},\\
&\{1,2,3,4,5\}\}\\
\end{array}
\begin{array}{rl}
\mathcal{A}_2=&\{\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\\
&\{1,2,3,4\},\{1,2,3,5\},\{1,2,4,5\},\\
&\{1,3,4,5\},\{2,3,4,5\},\\
&\{1,2,3,4,5\}\}\\
\end{array}
\end{align*}
Example: $n=5,k=3$
$|\mathcal{A}_1|=2^{5-3}=4$, $|\mathcal{A}_2|=M(5,3)=6$
\begin{align*}
\begin{array}{rl}
\mathcal{A}_1=&\{\{1,2,3\},\\
&\{1,2,3,4\},\{1,2,3,5\},\\
&\{1,2,3,4,5\}\}\\
\end{array}
\begin{array}{rl}
\mathcal{A}_2=&\{\{1,2,3,4\},\{1,2,3,5\},\{1,2,4,5\},\\
&\{1,3,4,5\},\{2,3,4,5\},\\
&\{1,2,3,4,5\}\}\\
\end{array}
\end{align*}
Example: $n=6,k=2$
$|\mathcal{A}_1|=2^{6-2}=16$, $|\mathcal{A}_2|=M(6,2)=22$
\begin{align*}
\mathcal{A}_1=\{&\{1,2\},\\
&\{1,2,3\},\{1,2,4\},\{1,2,5\},\{1,2,6\},\\
&\{1,2,3,4\},\{1,2,3,5\},\{1,2,3,6\},\\
&\{1,2,4,5\},\{1,2,4,6\},\{1,2,5,6\}\\
&\{1,2,3,4,5\},\{1,2,3,4,6\},\{1,2,3,5,6\},\{1,2,4,5,6\},\\
&\{1,2,3,4,5,6\}\}\\
\\
\mathcal{A}_2=\{&\{1,2,3,4\},\{1,2,3,5\},\{1,2,3,6\},\{1,2,4,5\},\{1,2,4,6\},\\
&\{1,2,5,6\},\{1,3,4,5\},\{1,3,4,6\},\{1,3,5,6\},\{1,4,5,6\},\\
&\{2,3,4,5\},\{2,3,4,6\},\{2,3,5,6\},\{2,4,5,6\},\{3,4,5,6\},\\
&\{1,2,3,4,5\},\{1,2,3,4,6\},\{1,2,3,5,6\},\\
&\{1,2,4,5,6\},\{1,3,4,5,6\},\{2,3,4,5,6\}\\
&\{1,2,3,4,5,6\}\}\\
\end{align*}
Best Answer
My answer is inspired by Ewan Delanoy’s that. He bounded size of $\mathcal F$ in terms of Ramsey numbers, showing that $|{\mathcal F}| \leq 2+R(2,R(4,4,3))=2+ R(4,4,3)$. The best known bounds for $R(4,4,3)$ are $55\le R(4,4,3)\le 77$, see [R, p. 39]. We shall show that $|\mathcal F|\le 14$.
Following Ewan Delanoy, given a subset $A$ of $[n]$ we put $A^+=A$, $A^-=A^c$ and let $A^{\pm}$ denotes $A$ or $A^c$. Subsets $A$ and $B$ of $[n]$ are independent, if all four intersections $A^{\pm}\cap B^{\pm}$ are nonempty. Assume that $|\mathcal F|\ge 3$.
Lemma. The family $\mathcal F$ does not contain independent members.
Proof. Suppose to the contrary that $A, B\in\mathcal F$ are independent and $C$ be an arbitrary member of $C$ distinct from $A$ and $B$. For any choice $^*, ^{**}$ of signs in $\pm$, a set $A^*\cap B^{**}$ is nonempty, so at least one of sets $A^*\cap B^{**}\cap C^+$ and $A^*\cap B^{**}\cap C^-$ is non-empty. So the family of sets of the form $A^\pm\cap B^\pm\cap C^\pm$ has at least four non-empty members, a contradiction. $\square$
Let a family $\mathcal F^*$ is obtained from family $\mathcal F\setminus\{\varnothing, [n]\} $ by replacing each member $A$ of $\mathcal F$ with $|A|>n/2$ by $A^c$. Then $|\mathcal F^*|\ge |{\mathcal F}|/2-1$ and $\mathcal F^*$ satisfies the question condition. Since $A^c\cap B^c$ is non-empty for each $A,B\in\mathcal F^*$, Lemma implies that any members of $\mathcal F^*$ are disjoint or one is contained in the other. It follows that any member of $\mathcal F^*$ contains a minimal element and minimal elements of $\mathcal F^*$ are pairwise disjoint. If $\mathcal F^*$ contains four minimal members $A$, $B$, $C$, and $D$ then sets $A\cap B^c\cap C^c=A$, $A^c\cap B\cap C^c=B$, $A^c\cap B^c\cap C=C$, and $A^c\cap B^c\cap C^c\supset D$ are non-empty, a contradiction. Thus $\mathcal F$ contains at most three minimal elements. Let $A$ be any of them. Suppose to the contrary that there exist distinct elements $B\supset A$ and $C\supset A$ of $\mathcal F^*\setminus \{A\}$. Since $B\cap C\supset A$ is non-empty, it follows that $B\subset C$ or $C\subset B$. Anyway, sets $A^c\cap B^c\cap C^c$, $A^c\cap B^c\cap C$, $A^c\cap B\cap C$, and $A\cap B\cap C$ are non-empty, a contradiction. Thus each minimal member of $\mathcal F^*$ is contained in at most one other member. Thus $|\mathcal F^*|\le 3\cdot 2=6$, and $|\mathcal F|\le 2(|\mathcal F^*|+1)\le 14$.
References
[R] Stanisław P. Radziszowski, Small Ramsey numbers. Dynamic Surveys. Electronic Journal of Combinatorics, revision #15: March 3, 2017.