The maximal number of antichains in a connected poset on $n$ elements is $2^{n-1}+1$, if you count $\emptyset$ as an antichain. 

It is achieved by the poset $Q(n)$ consisting of a single minimal element and an $(n-1)$-element antichain, each element of the antichain greater than the unique minimum.

To show that one cannot do any better consider the upper bound for the number of antichains in a poset (mentioned in [this answer][1]): If $P$ can be partitioned into $a$ disjoint chains and those chains consist of $c_1,c_2,\ldots,c_a$ elements, then the number of antichains in $P$ is at most $(c_1+1)(c_2+1)\cdots (c_a+1)$.

Thus, if there is a $3$-element chain in $P$, then the number of antichains is at most $$4\cdot 2^{n-3}=2^{n-1}<2^{n-1}+1.$$ 
**EDIT:**<br> *The following is wrong. Thanks to Seb Destercke for noting it.*<br>
<del>If there are two disjoint $2$-element chains, then it is at most $$\not 3 \not\cdot \not 3 \not\cdot \not 2^{\not n\not -\not 4}\not <\not 2^{n-1}\not+\not 1.$$</del>

*Corrected version:*<br>
If there are three disjoint $2$-element chains, then the number of antichains is at most $$3\cdot 3\cdot 3 \cdot 2^{n-6}=(2^5-5)\cdot 2^{n-6}=2^{n-1}-5\cdot 2^{n-6}<2^{n-1}+1.$$ 

It is easy to see that a connected poset without $3$-element chains and such that every pair of $2$-element chains has nonempty intersection must be $Q(n)$ or its dual. A connected poset without $3$-element chains that has two, but not three disjoint $2$-element chains is, up to duality, one of the following two kinds of posets, plus possibly some extra comparabilities (which decrease the number of antichains).

              [k]  B [n-k-3]
    X(n,k):     \ / \ /
                 A   C
    

              [k]  B
    Y(n,k):      \ | \
                   A  [n-k-2]

Here $A, B, C$ are vertices and $[m]$ denotes an $m$-element antichain. We need to show that $X(n,k)$ and $Y(n,k)$ both have at most as many antichains as $Q(n)$.

We may assume that the partial orders $X(n,k), Y(n,k)$ and $Q(n)$ are all defined on the same $n$-element base set and moreover $A\in Q(n)$ is the unique minimal element (thus $B,C\in Q(n)$ are maximal).

              B  C [n-3]
    Q(n):      \ | /
                 A 

The following defines an injective map from antichains in $X(n,k)$ or antichains in $Y(n,k)$ to antichains in $Q(n)$:
$$ \{x_1,\ldots,x_m\}\mapsto \{x_1,\ldots,x_m\} \quad \text{ if } x_i\not=A \text{ for all } 1\leq i\leq m,\\\{A\}\mapsto \{A\},\\ \{A,y_1,\ldots,y_m\}\mapsto \{B,y_1,\ldots,y_m\}.$$
Thus $X(n,k)$ and $Y(n,k)$ have at most as many antichains as $Q(n)$.

I believe one can find a more elegant proof (and would be happy to see one!). On the other hand, I think this question seems more appropriate for [math.SE][2].


  [1]: http://mathoverflow.net/questions/49515/on-the-number-of-antichains-of-a-poset/49550#49550
  [2]: http://math.stackexchange.com/