This is Exercise 3 of Chapter 3 in G.P. Tolstov's Fourer Series (1976).
If $P\left(x\right)=a_0+a_{1}\left(x\right)+\cdots+a_{n}x^{n}$ is a polynomial satisfying $\sum_{i=0}^{n} a_{i}^2=1$, then $$\int_{0}^{1}\vert P\left(x\right)\vert\text{d}x\le\frac{\pi}{2}.$$ [Show that the inequality continues to hold of the upper bound is replaced with $\pi/\sqrt{6}$.]
Up to this point, Bessel's inequality and Schwarz's inequality have been provided. My work is provided below, but I am skeptical because (i) my upper bound does not involve $\pi$ and (ii) my upper bound of $1$ is better than $\pi/\sqrt{6} \approx 1.28$. Below is the work followed by the reasoning. Why does $\pi$ appear in the author's inequality? \begin{align*} \int_{0}^{1}\left|P\left(x\right)\right|\text{d}x & =\int_{0}^{1}\left|a_{0}+a_{1}x+\cdots+a_{n}x^{n}\right|\text{d}x\\ & \le\int_{0}^{1}\left(\left|a_{0}\right|+\cdots+\left|a_{n}\right|\right)\\ & \le\int_{0}^{1}\left(\left|a_{0}\right|+\cdots+\left|a_{n}\right|\right)\left(1+0+\cdots+0\right)\text{d}x\\ & \le\int_{0}^{1}\sqrt{\left(\left|a_{0}\right|^{2}+\cdots+\left|a_{n}\right|^{2}\right)\left(1^{2}+0^{2}+\cdots+0^{2}\right)}\\ & =\int_{0}^{1}1\text{dx}\\ & =1. \end{align*}
The first inequality is due to the triangle inequality and fact that the interval of integration is between $-1$ and $1$, and the third inequality is due to the Cauchy-Schwarz inequality. The second-to-last equation follows from the assumption that the sum of the squares of the $a_{i}$'s is $1$.
By hypothesis, we have $$ a_0^2+\cdots+a_n^2=1 $$ As noted in the comments, your error is your assertion that $$ |a_0|+\cdots+|a_n| \le \sqrt{|a_0|^2+\cdots+|a_n|^2} $$ follows from the Cauchy-Schwarz inequality.
As a simple counterexample, for $n=1$, letting $a_0={\large{\frac{4}{5}}}$ and $a_1={\large{\frac{3}{5}}}$, we get $$ |a_0|+|a_1| = \frac{7}{5} > 1 = \sqrt{|a_0|^2+|a_1|^2} $$ and also $$ \left|\int_0^1 a_0+a_1x\right|=\frac{11}{10} > 1 $$ As regards the exercise in question, it can be resolved as follows . . . \begin{align*} & \left|\int_0^1 \sum_{k=0}^n a_kx^k\right| \\[4pt] \le\;& \int_0^1 \sum_{k=0}^n |a_k|x^k \\[4pt] =\;& \sum_{k=0}^n \frac{|a_k|}{k+1} \\[4pt] \le\;& \left(\sqrt{\sum_{k=0}^n a_k^2}\right) \left(\sqrt{\sum_{k=0}^n \frac{1}{(k+1)^2}}\right) \\[4pt] =\;& \sqrt{\sum_{k=0}^n \frac{1}{(k+1)^2}} \\[4pt] =\;& \sqrt{\sum_{k=1}^{n+1} \frac{1}{k^2}} \\[4pt] < \;& \sqrt{\sum_{k=1}^\infty \frac{1}{k^2}} \\[4pt] =\;& \sqrt{\frac{\pi^2}{6}} \\[4pt] =\;& \frac{\pi}{\sqrt{6}} \\[4pt] < \;& \frac{\pi}{2} \\[4pt] \end{align*}