I'll replace $|\sin t/2|$ by $|t|$ since they are comparable: $\frac{|t|}{\pi}\le \left|\sin\frac{t}{2}\right| \le \frac{|t|}{2} \ \text{ for }t\in [-\pi,\pi]$ Claim: for all $\lambda\ge 1$ $\frac{1}{3}\log \lambda \le \int_0^\pi \frac{|\sin \lambda t|}{t}\,dt \le \log \lambda +\log \pi +1.$
Proof. For the upper bound, split the integral into "small $t$" part and the rest: $\int_0^{\lambda^{-1}} \frac{|\sin \lambda t|}{t}\,dt \le \int_0^{\lambda^{-1}} \frac{\lambda t}{t}\,dt =1$ and $\int_{\lambda^{-1}}^\pi \frac{|\sin \lambda t|}{t}\,dt \le \int_{\lambda^{-1}}^\pi \frac{1}{t}\,dt = \log\lambda + \log \pi$
The lower bound needs a bit more work. Since the integrand is nonnegative, we can restrict the region of integration to the set $|\sin \lambda t|\ge 1/2$. This set contains the intervals $I_k=[\pi \lambda^{-1} (k+1/6), \pi \lambda^{-1} (k+5/6)]$ for all integers $k$ such that $0\le k \le \lambda-1$. The integral over $I_k$ is at least $ \int_{I_k} \frac{1/2}{t}\,dt \ge |I_k| \frac{1/2}{\pi \lambda^{-1} (k+1)} = \frac{1/3}{k+1} $ Therefore, the integral is bounded from below by $\frac13 \sum_{k=0}^{\lfloor \lambda-1\rfloor }\frac{1}{k+1} \ge \frac{1}{3} \log \lambda.$
Remark. If $\|D_n\|_{L^1}=o(\log n)$ were true, the estimate in #19 would hold for the Fourier series of any finite measure on $[-\pi,\pi]$. This is not the case. The fact that $s_n$ comes from a continuous function should be used.