7. By the property of absolute value, replacing $x$ with $-x$ does not change this relation. This means that the figure defined by the given inequality is symmetric with respect to the OY axis. Therefore, it is sufficient to find the area of half of the figure for $x \geq 0$. In this case, we obtain the inequality $\left|x-2 y^{2}\right|+x+2 y^{2} \leq 8-4 y$. By removing the absolute value sign, we get two regions: 1: $\left\{\begin{array}{c}x \geq 2 y^{2} \\ x-2 y^{2}+x+2 y^{2} \leq 8-4 y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 2 y^{2} \\ y \leq 2-\frac{x}{2}\end{array}\right.\right.$

Region II: $\left\{\begin{array}{c}x \leq 2 y^{2} \\ -x+2 y^{2}+x+2 y^{2} \leq 8-4 y\end{array} \Leftrightarrow\left\{\begin{array}{c}x \leq 2 y^{2} \\ y^{2}+y-2 \leq 0\end{array} \Leftrightarrow\left\{\begin{array}{c}x \leq 2 y^{2} \\ -2 \leq y \leq 1\end{array}\right.\right.\right.$

Next, on the coordinate plane xOy, we plot the graphs of the obtained inequalities for $x \geq 0$, taking into account that $x=2 y^{2}$ is the graph of a parabola with its vertex at the origin, and the branches of this parabola are directed along the positive direction of the Ox axis. The union of these regions gives a figure which is a trapezoid $M N A B$, the area of which is equal to $S=\frac{1}{2} * 3(8+2)=15$. Then the doubled area is 30 (see Fig. 2).