Suppose $E=A+B$ is the union of two sets $A$ and $B$.Then, we can have $4$ ways of partitioning $E$ into disjoint subsets: $\{AB,A∆B\},\{B, A\bar B\},\{A,\bar A B \},\{AB,A\bar B,\bar AB\}$ which is straightforward to identify from the Venn diagram as non-overlapping regions, the smallest such regions being $AB,A\bar B $ and $ \bar AB\ $. However I am unable to get any clue on how to generalize this to $3$ or more sets (becaus eit gets very clumsy to use Venn diagrams for the purpose): in case of $3$ sets there are $7$ smallest possible disjoint subsets. I guess there should be $\geq 8$ ways of partitioning this set. In fact there should be $2^n-1$ smallest possible disjoint subsets of the union $E_n=A_1+A_2+...+A_n$ and there should be, in fact the $\geq 2^n$ partitions possible for $E_n$. However, this is just a guess, I believe that this could be wrong. But i couldn't find anything relating to this on the net or any of my textbooks, except that it is related to Boolean factorisation and consensus sets. So, any reference would be helpful.
Edit
The number of such partitions must be more than $2^n$. For the case $n=3$ here are some of the desired partitions:
Update $-1$
As I understand $P=\{E_1, E_2, ..., E_l \}$ is a partition of $E$ iff all the $E_i$'s are non-empty, $E_iE_j=\emptyset$ for all pairs $i\neq j$ and $E_1 + E_2 + ... + E_l = E$. Now, any set $E_{ij} = E_i + E_j$ will still be disjoint with the remaining $l-2$ subsets of $E$ and hence forms another partition with them. In this way there are $\text{C}(l,2)$ ways to form partitions of size $l-1$ from $P$ viz. $\{ E_1,E_2,...,E_{i-1},E_{i+1},...,E_{j-1},E_{j+1},...,E_l \}$. Applying the formula recursively we find the total number of partitions of decreasing size to be $\text{C}(l,2)\text{C}(l-1,2)...\text{C}(2,2) + 1 = \frac{l!(l-1)!}{2^{l-1}}+1$. But for an $n$-union set $E=E_n$, $l=2^n -1$. In this way, the desired number is $56701$ for $n=3$ and $6958057668962400001$ for $n=4$, much larger than what @joriki has answered.
Update $-2$
Seems my above analysis is partly wrong. For $P = \{ E_1,E_2,E_3,E_4\}$, the different partitions are: $1$ partition of type $\{E_i,E_j,E_k,E_l\}$ corresponding to $(1+1+1+1)$, $\text{C}(4,2)=6$ partitions of type $\{E_i+E_j,E_k,E_l\}$ corresponding to $(2+1+1)$, $\text{C}(4,2)/2=3$ partitions of type $\{E_i+E_j,E_k+E_l\}$ corresponding to $(2+2)$,$\text{C}(4,3)=4$ partitions of type $\{E_i+E_j+E_k,E_l\}$ corresponding to $(3+1)$ and the total number is $14$. Now the number for some $P=\{E_1,...,E_{2^n -1}\}$ is desired, which is exactly $B_{2^n-1}-1$.
I assume that you want to count the (equivalence classes of) expressions that can be used to partition general sets, irrespective of whether any of the resulting subsets are actually non-empty for any particular sets, i.e. whether these formally distinct expressions actually correspond to distinct partitions of any particular sets. However, judging from your example and your count of $2^n-1$, you do want to exclude expressions like $\bar A\bar B$ that yield empty subsets for all $A$ and $B$. Also, contrary to convention, you’re apparently not counting $\{E\}$ as a partition of $E$.
As you say, given $n$ sets $A_n$, there are $2^n-1$ different products of the sets and their complements, since for each set you can include either the set or its complement, yielding $n$ binary choices, and the product with only complements is to be excluded because it yields an empty set for any $A_n$.
Every formal partition expression can be expressed as a partition of these products. The number of partitions of a set with $k$ labeled elements is given by the Bell number $B_k$. We need to subtract one again because you’re not counting $\{E\}$ as a partition of $E$. Thus, the number of (equivalence classes of) expressions that can be used to partition a union of $n$ sets in the desired sense is $B_{2^n-1}-1$. Here’s a table of the first few values:
\begin{array}{r|r} n&B_{2^n-1}-1\\\hline 1&0 \\ 2&4 \\ 3&876 \\ 4&1382958544 \\ 5&10293358946226376485095652 \end{array}
The Bell numbers grow very roughly as $n^n$ (see the Wikipedia article for more precise asymptotics), so your counts grow very roughly as $\left(2^n\right)^{2^n}=2^{n\cdot2^n}$.