As a general comment on this kind of construction, I might suggest reading this answer.
Generally speaking, what you describe is the "top-down" approach to the construction. It is not, generally, practical to actually know the description of the elements in the generated object. For that, you want the "bottoms-up" construction. In the question linked-to above, Asaf Kargila provides the description of the bottoms-up construction.
In your particular case, of course, since $E$ is small (3 elements), so $P(E)$ is small (8 elements), the number of possible $\sigma$-algebras is somewhat manageable (though still large).
What are all $\sigma$-algebras on $E=\{1,2,3\}$ that contain $\{1\}$ as an element? They must contain $\emptyset$, $\{1\}$, $\{2,3\}$, and $E$. There are 4 other elements in $P(E)$ which may or may not be in a $\sigma$-algebra.
- One $\sigma$-algebra is just $\{\emptyset, \{1\}, \{2,3\}, E\}$.
- If the $\sigma$-algebra contains either $\{2\}$ or $\{3\}$ in addition to those, then it must also contain the other (symmetric difference with $\{2,3\}$), and hence be all of $P(E)$.
- If the $\sigma$-algebra contains any other $2$-element subset, then it must contain another singleton (symmetric difference with $\{2,3\}$), hence must be all of $P(E)$.
So in fact the only $\sigma$-algebras on $E$ that contain $\{1\}$ are $\{\emptyset, \{1\},\{2,3\}, E\}$ and $P(E)$. The intersection is just $\{\emptyset,\{1\},\{2,3\},E\}$, so the $\sigma$-algebra generated by $\{1\}$ is $\{\emptyset, \{1\}, \{2,3\}, E\}$.
Consider set $E_1,\dots,E_N$. Take $S_I:=\bigcap_{i\in I}E_i\cap\bigcap_{i\in [N]\setminus I}E_k^c$ for $I\subset [N]$. These sets are pairwise disjoint and generate the same $\sigma$-algebra as $E_1,\dots,E_N$. For each $I\subset [2^N]$, define $F_I:=\bigcup_{i\in I}S_i$. Then the $\sigma$-algebra generated by the $E_j$ is $\{F_I,I\subset [N+1]\}$.
Note that the $\sigma$-algebra generated by a finite collection is finite.
Best Answer
Here is a method to compute the $\sigma$-algebra generated by two non-empty disjoint sets say $S_1$ and $S_2$ on a set $S$.
This extends readily to the $\sigma$-algebra generated by a finite collection of non-empty pairwise disjoint subsets $S_1,\dots,S_n$: let $S_{n+1}:= S\setminus \bigcup_{i=1}^nS_i$: then $$ \sigma(S_i,1\leqslant i\leqslant n)=\left\{ \bigcup_{q=1}^n S_q^{i_q},i_q\in \{0,1\} \right\}. $$