To get the actual answer out of the way: the usual definition of ideal implies that any ideal contains the empty set -- an **ideal** $I$ (on a set $X$ / the power set of $X$) is non-empty, closed under taking subsets and under taking finite unions (and of course $I\subseteq P(X)$). The first two should convince you that it contains the empty set. 

The notion of filter is dual -- non-empty, closed under taking finite intersections and supersets. Ideals correspond to filters by mapping each element $A\in I$ to its complement $X\setminus A$. Maximal ideals correspond to maximal filters.


Just in case this question was only due to a confusion of ideals and filters, let me add:

A **proper ideal** by definition does not contain the 'full' set $X$ (e.g. in your example $X = \{ 1,2,3 \}$). Similarly, a proper filter does not contain the empty set by definition. 
The **'improper'** cases of these definitions coincide -- both the improper filter and the improper ideal are just the full power set (as is clear from being closed under subsets and supersets respectively).

Usually, **filter/ideal means _proper_ filter/ideal**, but for notational or technical convenience, it sometimes seems nice to allow the improper case -- for example in the Stone-Cech compactification of the natural numbers, $\beta \mathbb{N}$, proper filters (on $\mathbb{N}$) correspond to closed non-empty subsets and the improper filter to the empty set. But I think the general preference (as Joel David Hamkins pointed out in the comments) is not to do this since no convenience outweighs the **confusion caused by the improper case**.

In your example and (as mentioned by Robin Chapman) for any **finite set** $X$ , the maximal (proper) filters (or ultrafilters) are the **principal** filters, i.e. those of the form 
$\dot{x} = \{ A \subseteq X:\ x \in A \} $ for some $x\in X$. To see this just partition $X$ into singletons -- a finite partition by assumption on $X$ -- every maximal filter contains exactly one part of the partition. The maximal ideals are again the dual.