I don't think so. Each generator $l$ of $L$ is a vector of length $2^n$ with each entry $\pm a^i$ for some $0\le i \le n.$ There is only one entry which is $a^n$. If $a$ is large enough, then wouldn't that $k=1$ entry be larger than $\|l\| \sqrt 1 / \log\log n ? $

I don't think so. Each generator $l$ of $L$ is a vector of length $2^n$ with $\binom{n}{i}$ entries equal to $\pm a^i$ for $0\le i \le n.$ So $\|l\|=(a^2+1)^{n/2}$ and there is only one entry which is $a^n$. If $a$ is large enough, then wouldn't that $k=1$ entry be larger than $\|l\| \sqrt 1 / \log\log n ? $