Yes. Denote $\alpha = \sqrt{2} + \sqrt{3} + \sqrt{5}$. Then $\alpha^{2k-1} = \lambda_{1,k}\sqrt{2} + \lambda_{2,k}\sqrt{3} + \lambda_{3,k}\sqrt{5} + \lambda_{4,k}\sqrt{30}$ for $k\in \{1,2,3,4\}$ and some integers $\lambda_{i,j}$, $1\leq i,j \leq 4$. Since $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{30}$ are linearly independent over $\mathbb{Q}$, the system has non-zero determinant, and so $\sqrt{2} = \dfrac {\mu_{1}\alpha + \mu_{2}\alpha^3 + \mu_{3}\alpha^5 + \mu_{4}\alpha^7} {\mu}$ for some integers $\mu_k$, $1\leq k\leq 4$ and $\mu \neq 0$. Therefore $\sqrt{2}$ is constructible, and so $1$ is constructible. A more direct approach is using the fact that $\alpha$ is a primitive element of the degree $8$ algebraic extension of $\mathbb{Q}$.