jmc

We note that $4096=2^{12}$, so the expression becomes $$\sqrt{\sqrt{2^{12}}+\sqrt[3]{2^{12}}+\sqrt[4]{2^{12}}}=\sqrt{2^6+2^4+2^3}=\sqrt{64+16+8}=\sqrt{88}$$

Factoring out $4$ from $88$ gives $\boxed{2\sqrt{22}}$, which cannot be simplified further.