jmc

Setting $x=0,$ we get $|c|\le 1.$ Setting $x=1,$ we get $$|a+b+c|\le 1.$$Setting $x=\frac{1}{2},$ we get $$\left|\frac{a}{4}+\frac{b}{2}+c\right|\le 1.$$Let $$\begin{array}{rl}p& =c,\\ q& =\frac{a}{4}+\frac{b}{2}+c,\\ r& =a+b+c,\end{array}$$so $-1\le p,$ $q,$ $r\le 1.$ Solving for $a,$ $b,$ and $c,$ we find $$\begin{array}{rl}a& =2p-4q+2r,\\ b& =-3p+4q-r,\\ c& =p.\end{array}$$Hence, by Triangle Inequality, $$\begin{array}{rl}|a|& =|2p-4q+2r|\le |2p|+|4q|+|2r|=8,\\ |b|& =|-3p+4q-r|\le |3p|+|4q|+|r|=8,\\ |c|& =|p|\le 1.\end{array}$$Therefore, $|a|+|b|+|c|=8+8+1=17.$

Consider the quadratic $f(x)=8x^2-8x+1.$ We can write $$f(x)=8{\left(x-\frac{1}{2}\right)}^2-1.$$For $0\le x\le 1,$ $0\le {\left(x-\frac{1}{2}\right)}^2\le \frac{1}{4},$ so $-1\le f(x)\le 1.$

Therefore, the largest possible value of $|a|+|b|+|c|$ is $\boxed{17}.$