Given $\epsilon > 0$, what is the minimal $d$, such that $\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{d} \ge \epsilon 2^n$? Is there a clean answer?

There's no clean answer, but you need to use the Central Limit Theorem. If $X_1,\ldots, X_n$ are independent random variables taking values 0 and 1 with probability $1/2$, then you're asking for a $d$ such that $\mathbb P(S_n\le d)\ge \epsilon$. By the Central limit theorem, for large $n$, you have $S_n$ is well approximated by a normal distribution with mean $n/2$ and variance $n/4$. That is $S_n$ has approximately the same distribution as $n/2+\sqrt n/2 N$ where $N$ is a Normal(0,1) random variable. In this language, you are now asking for $d$ such that $\mathbb P(N\le (2dn)/\sqrt n)\approx\epsilon$, or $\Phi((2dn)/\sqrt n)\approx\epsilon$. If you solve $\Phi(x)=\epsilon$, then you can obtain a suitable value for $d$ is something like $n/2\sqrt{2n\log\epsilon}$. 

