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
If I understand your question correctly, you can't do better. Indeed from the proof of Erdos-Ko-Rado you can deduce that only the stars have size equal to $\binom{n-1}{r-1}$, when $2r<n$. In the exceptional case $r=n/2$ you can take one of every pair of complementary sets $A,A^c$.
EDIT: The intended question seems to be (equivalent to) the following. How large can an intersecting family of $r$-sets be if we stipulate that it is not a star?
The answer in this case was given by Hilton and Milner (see A. J. W. Hilton and E. C. Milner. Some Intersection Theorems For Systems of Finite Sets Q J Math (1967) 18(1): 369-384 doi:10.1093/qmath/18.1.369).
The largest such family is obtained by fixing an element $1$, an $r$-set $A_1 = \{2,\dots,r+1\}$, and requiring every further set to contain $1$ and to intersect $A_1$. Thus the maximum size is whatever the size of this family is.
Since you seem to be particularly interested in the case of $n$ odd and $r=\lfloor n/2\rfloor$, let's do the calculation in that case. The family you propose ("exclude two points") has size $\binom{n-2}{r} = \binom{n-2}{r-1}$, while the family that Hilton and Milner propose has size $$1 + \binom{n-1}{r-1} - \binom{n-r-1}{r-1} = \binom{n-1}{r-1} - r + 1 = \binom{n-2}{r-1} + \binom{2r-1}{r-2} - r + 1.$$
Sorry, they win. :)