I claim the following result which implies what you are asking for.
Claim: Let $g$ be a convex function such that $\int_0^1 g=0$. Then one of $\int_0^{1/4}g$ and $\int_{3/4}^1 g$ is at least $\|g\|_1/18$.
The result implies what you are looking for (up to replacing 8 by 9) by setting $g=1-f$.
Proof: Since $\int_0^1 g=0$, we have $\int g^+=\int g^-=\|g\|_1/2$. Let $\{x\colon g(x)\le 0\}=[a,b]$.
First suppose that $b\le\frac 34$. In this case, since $g$ is increasing on $[b,1]$, $\int_{3/4}^1 g\ge \int_b^{b+\frac 14}g$. By convexity, $\int_b^{b+1/4}g\ge \frac1{32}g'(b^+)$ while $\int g^-=-\int_a^b g\le \frac 12(b-a)^2g'(b^+)\le \frac{9}{32}g'(b^+)$. It follows that $\int_{3/4}^1 g\ge \frac 19\int g^-=\frac 1{18}\|g\|_1$. By symmetry, the same applies if $a\ge \frac 14$.
In the other case, $g$ is negative on $[\frac 14,\frac 34]$. Since $\int g=0$, either $\int_0^{1/2}g\ge 0$ or $\int_{1/2}^1 g\ge0$. Without loss of generality, we assume that the first of these holds. Hence $\int_0^{1/4}g\ge -\int_{1/4}^{1/2}g$. We then apply the following lemma with $f=-g$, which gives $\int_{1/4}^{1/2}(-g)\ge \frac 19\int (-g)^+$, or $-\int_{1/4}^{1/2}g\ge \frac 1{18}\int\|g\|_1$, and the claim then follows.
Lemma: Let $f$ be concave on $[\frac 14,1]$ and positive on $[\frac 14,\frac 34]$ (at least). Then $\int_{1/4}^{1/2}f\ge \frac 19\int f^+$.
Proof: We are attempting to find a lower bound for $\int_{1/4}^{1/2}f/ \int_0^1 f^+$. Given any $f$, this quantity is reduced if $f$ is replaced on $[\frac 14,\frac 12]$ by a linear function joining $(\frac 14,f(\frac 14))$ to $(\frac 12,f(\frac 12))$. It is then further reduced by extending that linear function to $[\frac 14,1]$. Now it is straightforward to see that the quantity is minimized if $f(x)=x-\frac 14$ on $[\frac 14,1]$, where the ratio is $\frac 19$ as required.
