3. Given the quadratic function $y=x^{2}+a x+b$ whose graph intersects the $x$-axis at two points with the x-coordinates $m$ and $n$, and $|m|+|n| \leqslant 1$. Let the maximum and minimum values of $b$ that satisfy the above conditions be $p$ and $q$, respectively. Then $|p|+|q|$
3. $\frac{1}{2}$.

According to the problem, $m$ and $n$ are the roots of the quadratic equation $x^{2} + a x + b = 0$, then
$$
\begin{array}{l}
m+n=-a, \quad m n=b. \\
\text { Also, } |m|+|n| \leqslant 1, \text { then } \\
|m+n| \leqslant |m|+|n| \leqslant 1, \\
|m-n| \leqslant |m|+|n| \leqslant 1.
\end{array}
$$

Since the discriminant of the equation $x^{2} + a x + b = 0$ is
$$
\Delta = a^{2} - 4 b \geqslant 0,
$$

we have $b \leqslant \frac{a^{2}}{4} = \frac{(m+n)^{2}}{4} \leqslant \frac{1}{4}$.
$$
\begin{array}{l}
\text { Therefore, } 4 b = 4 m n = (m+n)^{2} - (m-n)^{2} \\
\quad \geqslant (m+n)^{2} - 1 \geqslant -1 \\
\Rightarrow b \geqslant -\frac{1}{4},
\end{array}
$$

with equality holding if and only if $m = -n = \frac{1}{2}$;
$$
\begin{aligned}
4 b & = 4 m n = (m+n)^{2} - (m-n)^{2} \\
& \leqslant 1 - (m-n)^{2} \leqslant 1 \\
\Rightarrow b & \leqslant \frac{1}{4},
\end{aligned}
$$

with equality holding if and only if $m = n = \frac{1}{2}$.
Thus, $p = \frac{1}{4}, q = \frac{1}{4}$.
Therefore, $|p| + |q| = \frac{1}{2}$.