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Taking the absolute value of both sides, we get $$|z^3|=|2+2i|=2\sqrt{2}.$$Then $|z{|}^3=2\sqrt{2},$ so $|z|=\sqrt{2}.$

Let $w=\frac{z+\overline{z}}{2},$ so the possible values of $w$ are $a_1,$ $a_2,$ and $a_3.$ Then $$w^3=\frac{z^3+3z^2\overline{z}+3z{\overline{z}}^2+{\overline{z}}^3}{8}.$$We know that $z^3=2+2i.$ Taking the conjugate, we get $\overline{z^3}=\overline{2+2i},$ so ${\overline{z}}^3=2-2i.$ Also, $$3z^2\overline{z}+3z\overline{z}=3z\overline{z}(z+\overline{z})=6|z{|}^{2}w=12w,$$so $$w^3=\frac{2+2i+12w+2-2i}{8}=\frac{4+12w}{8}=\frac{3}{2}w+\frac{1}{2}.$$Then $$w^3-\frac{3}{2}w-\frac{1}{2}=0.$$By Vieta's formulas, $a_1{a}_2{a}_3=\boxed{\frac{1}{2}}.$