7. If real numbers $a, b, c$ make the quadratic function $f(x) = a x^{2} + b x + c$ such that when $0 \leqslant x \leqslant 1$, always $|f(x)| \leqslant 1$. Then the maximum value of $|a| + |b| + |c|$ is $\qquad$.
7. 17.

Take $x=0, \frac{1}{2}, 1$.
From the problem, we have
$$
\begin{array}{l}
|c| \leqslant 1,|a+2 b+4 c| \leqslant 4, \\
|a+b+c| \leqslant 1 .
\end{array}
$$

Let $m=a+2 b+4 c, n=a+b+c$.
Then $a=-m+2 n+2 c, b=m-n-3 c$
$$
\begin{aligned}
\Rightarrow & |a| \leqslant|m|+2|n|+2|c| \leqslant 8, \\
& |b| \leqslant|m|+|n|+3|c| \leqslant 8 \\
\Rightarrow & |a|+|b|+|c| \leqslant 17 .
\end{aligned}
$$

It is easy to verify,
$$
f(x)=8\left(x-\frac{1}{2}\right)^{2}-1=8 x^{2}-8 x+1,
$$

When $0 \leqslant x \leqslant 1$, it always holds that $|f(x)| \leqslant 1$. Therefore, the maximum value sought is 17.