I am looking for an efficient lower bound of the expression
$$ E(p)=\sum_{i=0}^{\left \lceil \frac{l}{2}\right \rceil-1}\sum_{j=0}^{i}\binom{l}{i}\binom{i}{j}\left ( -1 \right )^{j}p^{l-i+j}, $$ where $\frac{1}{2}<p<1$.
EDIT: As I suggested in comments, the expression should converge to $1$. I am very much interested in a lower-bound approximation, in terms of $p$ (you may assume that $l$ is large, and of any parity).
Ultimately, I would be interested in a rate of convergence of the sequence $p=p_0,p_1,\dots $ , where $ p_{i+1}=E(p_i)$.
I think there is no general answer. The speed of convergence depends on $l$ and $p$.
Be $\enspace \frac{1}{2}<p<E(p)<1$ .
Speed of convergence:
$E(1-p)=1-E(p)\leq c(1-p)^m$ with $0<c<1$ and $m\geq 1$
For odd $l$ we get $E(p)|_{p\to \frac{1}{2}+0}\to \frac{1}{2}+0$ which means $c\to 1^-$ and $m\to 1^+$ or in words: Very weak convergence for $p$ near by $\frac{1}{2}$ . Much better is it with e.g. $p>\frac{2}{3}$ .
$E(1-p)=\sum\limits_{i=\lceil l/2\rceil}^l \binom{l}{i}p^{l-i}(1-p)^i \leq c(1-p)^m$
It remains the question what we can choose for $m$ so that
$\sum\limits_{i=\lceil l/2\rceil}^l \binom{l}{i}p^{l-i}(1-p)^{i-m} \leq c<1$
If $p$ is near by $1$ then we can choose $m$ with $m<\lceil l/2\rceil$ because the left side is very small and we can find $c\in(0;1)$ .
The condition is e.g. $\sum\limits_{i=\lceil l/2\rceil}^l \binom{l}{i}p^{l-i}(1-p)^{i-\lceil l/2\rceil} \leq \frac{c}{1-p}<\frac{1}{1-p}$ . In this case we have convergence of order $\lceil l/2\rceil-1$.