I'm curious to find out if the sum can be expressed in some known constants. What do you think about that? Is it feasible? Have you met it before?
$$2 \left(\text{Li}_2\left(2-\sqrt{2}\right)+\text{Li}_2\left(\frac{1}{2+\sqrt{2}}\right)\right)+\text{Li}_2\left(3-2 \sqrt{2}\right)$$
Euler's reflection identity states:
$$\operatorname{Li}_{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(1-z\right)}=\zeta{(2)}-\ln{\left(z\right)}\ln{\left(1-z\right)}.$$
Landen's dilogarithm identity states:
$$\operatorname{Li}_{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(\frac{z}{z-1}\right)}=-\frac12\ln^2{\left(1-z\right)};~~~\small{z\notin(1,\infty)}.$$
Setting $z=2-\sqrt{2}$ in Euler's identity, we have
$$\operatorname{Li}_{2}{\left(2-\sqrt{2}\right)}=\zeta{(2)}-\ln{\left(2-\sqrt{2}\right)}\ln{\left(\sqrt{2}-1\right)}-\operatorname{Li}_{2}{\left(\sqrt{2}-1\right)}.$$
Setting $z=\frac{1}{2+\sqrt{2}}$ in Landen's identity, we have
$$\operatorname{Li}_{2}{\left(\frac{1}{2+\sqrt{2}}\right)}=-\frac12\ln^2{\left(\frac{1}{\sqrt{2}}\right)}-\operatorname{Li}_{2}{\left(1-\sqrt{2}\right)}.$$
Thus,
$$\begin{align} I &=2\left[\operatorname{Li}_{2}{\left(2-\sqrt{2}\right)}+\operatorname{Li}_{2}{\left(\frac{1}{2+\sqrt{2}}\right)}\right]+\operatorname{Li}_{2}{\left(3-2\sqrt{2}\right)}\\ &=2\left[\zeta{(2)}-\ln{\left(2-\sqrt{2}\right)}\ln{\left(\sqrt{2}-1\right)}-\operatorname{Li}_{2}{\left(\sqrt{2}-1\right)}\right]\\ &~~~~~+2\left[-\frac12\ln^2{\left(\frac{1}{\sqrt{2}}\right)}-\operatorname{Li}_{2}{\left(1-\sqrt{2}\right)}\right]+\operatorname{Li}_{2}{\left(3-2\sqrt{2}\right)}\\ &=2\zeta{(2)}-2\ln{\left(2-\sqrt{2}\right)}\ln{\left(\sqrt{2}-1\right)}-\frac14\ln^2{(2)}\\ &~~~~~-2\left[\operatorname{Li}_{2}{\left(\sqrt{2}-1\right)}+\operatorname{Li}_{2}{\left(-\left(\sqrt{2}-1\right)\right)}\right]+\operatorname{Li}_{2}{\left(3-2\sqrt{2}\right)}\\ &=2\zeta{(2)}-2\ln{\left((\sqrt{2}-1)\sqrt{2}\right)}\ln{\left(\sqrt{2}-1\right)}-\frac14\ln^2{(2)}\\ &~~~~~-\operatorname{Li}_{2}{\left(\left(\sqrt{2}-1\right)^2\right)}+\operatorname{Li}_{2}{\left(3-2\sqrt{2}\right)}\\ &=2\zeta{(2)}-2\ln^2{\left(\sqrt{2}-1\right)}-\ln{(2)}\ln{\left(\sqrt{2}-1\right)}-\frac14\ln^2{(2)}.\blacksquare\\ \end{align}$$