Do you want to know about specific $n$ values or asymptotics?
In the event that $e_i=e_j,$ $|A| \le 2^{n}-2^{n-2}$ otherwise it seems highly likely that it is larger than that and, in any case, that $|A| \gt 2^{n-1}.$ The chance of that happeninghaving some $e_i=e_j$ is about $4.3\%$ for $n=10$ but $0.018\%$$1-\prod_{j=1}^{19}(1-\frac{j}{2^{20}}) \approx 0.018\%$ for $n=20.$
The chance of having a case of $e_i+e_j=e_k$ is larger but still goes to $0$ as $n$ increases. If that does happen then $|A| \le 2^n-2^{n-3}$
In $1000$ trials with $n=10$ There were $40$ cases of $e_i=e_j$ and of these $|A|$ ranged from $552$ to $762.$ It the other $960$ cases (with the $e_i$ distinct) the sums ranged from $628$ to $1012.$
In $100$ trials with $N=20$ there were, as expected, no cases of $e_i=e_j.$ The values of $|A|$ ranged between $2^{20}-2^{17.19}$ and $2^{20}-2^{16.02}.$
LATER If you just want a lower bound which grows faster than any polynomial then I will propose $\sqrt{2}^N=2^{N/2}.$ This is of course terrible since the few experiments always had $|A| \gt 2^{N-1}.$ But lets look at the probability that if we just look at $e_1,e_2,\cdots,e_m$ with $2m \le N$ we have all $2^m$ partial sums unique. If there are two sets $I,J$ with $\sum_Ie_i=\sum_Je_j$ then there are two disjoint sets $I,J$ with this property (delete the common elements from both the old $I$ and old $J$.) To get one of these potential disjoint pairs $I,J$ we would split the indices $1,2,\cdot,m$ into three classes:
used in $I$
used in $J$ and
used in neither.
There are $3^m$ ways to do this (actually less than half this since we can ignore any possibility with $I$ or $J$ empty and we don't care which is $I$ and which is $J$.) The probability that a particular split gives equal sums is less than $2^{-N}$ since the maximum element can be one of $2^N$ values but at best, just one of these makes that split work. So the probability that at least one split gives equal sums is less than $\frac{3^m}{2^N}.$ Putting $m=N/2$ gives the probability of this failure at $\left(\frac{\sqrt{3}}{2}\right)^N$ which goes to $0$ exponentially quickly.