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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): 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 $$4\cdot 2^{n-3}=2^{n-1}<2^{n-1}+1.$$ If there are two disjoint $2$-element chains, then it is at most $$3\cdot 3 \cdot 2^{n-4}<2^{n-1}+1.$$ 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.

I believe one can find an easier proof. On the other hand, I think this question seems more appropriate for math.SE.