You can apply Chernoff Bound - It states $\sum_{i > \lfloor \frac{n}{2} \rfloor} \binom{n}{i} p^i (1-p)^{n-i} \ge 1 - e^{-\frac{n}{2p}(p-\frac{1}{2})^2}$.
If you write the sum in reverse and use the symmetry of binomial coefficients, you obtain
$\sum_{i < \lceil \frac{n}{2} \rceil} \binom{n}{i} p^{n-i} (1-p)^{i} \ge 1 - e^{-\frac{n}{2p}(p-\frac{1}{2})^2}$ or $\sum_{i < \lceil \frac{n}{2} \rceil} \binom{n}{i} (\frac{1-p}{p})^{i} \ge p^{-n}(1 - e^{-\frac{n}{2p}(p-\frac{1}{2})^2}) $
Choosing $p \in (0,1)$ such that $\frac{1-p}{p} = \alpha$ (assuming $\alpha > 0$), i.e. $p=\frac{1}{1+\alpha}$, we obtain
$\sum_{i < \lceil \frac{n}{2} \rceil} \binom{n}{i} \alpha^{i} \ge (1+\alpha)^{n}(1 - e^{-\frac{n}{8}\frac{(1-\alpha)^2}{(1+\alpha)}})$ So you can take the following $C$ that depends on $n$ and $p$, at least in the case $n_0=\lfloor \frac{n}{2} \rfloor$: $C=1 - e^{-\frac{n}{8}\frac{(1-\alpha)^2}{(1+\alpha)}}$
EDIT: By choosing $\alpha = \frac{p}{1-p}$ you can get the following upper bound: $\sum_{i \le \lfloor \frac{n}{2} \rfloor} \binom{n}{i} \alpha^{i} \le (1+\alpha)^{n} e^{-\frac{n}{8}\frac{(1-\alpha)^2}{\alpha(1+\alpha)}}$
EDIT 2: In general, if $\alpha>0$ and we divide your inequality by $(1+\alpha)^n$ we obtain: $\sum_{k\le n_0} \binom{n}{k} (\frac{\alpha}{1+\alpha})^{k} (\frac{1}{1+\alpha})^{n-k} \ge c$, so we want to bound the probability that among $n$ coin tosses, we got "Head" no more than $n_0$ times, where the probability for head is $\frac{\alpha}{1+\alpha}$. Under the condition $\alpha < 1$, I can replace that "8" in the denominator by "2", see Hoeffding.