The Chernoff bound you need is given by $ Pr(X < (1-\delta)Ex[X]) \le \left( \frac{e^\delta}{(1-\delta)^{1-\delta}} \right)^{Ex[X]}. $
In your case, $Ex[X] = n/4$. Therefore, $ Pr(X < (1-\epsilon)n/8) \le Pr(X < n/8) \le Pr(X < (1/2)n/4) = Pr(X < (1 - 1/2)n/4), $ so $\delta = 1/2$.
This leads to $ Pr(X < (1-\epsilon)n/8) \le \left( \frac{e^{1/2}}{(1/2)^{1/2}} \right)^{n/4}. $
The expression on the right is easily reduced by algebraic manipulation. After this, you need to find an $n$ such that this expression is polynomially small.
This shows some of the power of Chernoff bounds, that is, that once you know $Ex[X]$, you can pick $\delta$ easily, and then the bound itself is given by a simple closed expression raised to $Ex[X]$. Let me repeat: all you need to know (if you have 0/1 independent Poisson Trials), is the expected value!
If you would like help with the algebraic manipulation, please comment as such.
NB + spoiler: Another formulation to memorize (cf. deathbed formulae :-) of the lower Chernoff bound is $ Pr(X < (1-\delta)Ex[X]) \le e^{-Ex[X]\delta^2/2}. $ In your case, this would become $ Pr(X < (1-\epsilon)n/8) \le e^{-(n/4)(1/2)^2/2} = e^{-n/32}. $ Solving for $n$ such that this expression is less than $N^{-c}$ gives $n \ge 64\log N$ (to get the bound with high probability with respect to $N$), as this equation shows: $ e^{-64\log(N)/32} = \left( e^{\log N} \right)^{-2} = N^{-2}. $
Thus, with at least $n = \Theta(\log N)$ flips (or whatever your variables denote) means that $X \ge (1-\epsilon)n/8$ with high probability with respect to $N$.