How many operations to produce this superposition of 3 qubits?
You don't need a full 8x8 three-qubit operation, but you will need a two-qubit operation of some kind. If you only rotate each qubit independently, you'll find you can only make un-entangled states like:

$(a_1 \left| 0 \right\rangle + b_1 \left| 1 \right\rangle) \otimes (a_2 \left| 0 \right\rangle + b_2 \left| 1 \right\rangle) \otimes (a_3 \left| 0 \right\rangle + b_3 \left| 1 \right\rangle)$

$= a_1 a_2 a_3 |000\rangle + a_1 a_2 b_3 |001\rangle + a_1 b_2 a_3 |010\rangle + a_1 b_2 b_3 |011\rangle + b_1 a_2 a_3 |100\rangle + b_1 a_2 b_3 |101\rangle + b_1 b_2 a_3 |110\rangle + b_1 b_2 b_3 |111\rangle$

But as soon as you have a non-trivial two-qubit operation, like a controlled-NOT, to go with your single qubit gates, you can approximate any larger operation or state arbitrarily well. (This is different from classical reversible computing, where you need a doubly-controlled-not before you get universality.)

The basic idea of how it's done is:


Break a large operation into lots of single-qubit operations, but allowing each to have lots of controls.
Use the fact that all single-qubit operations have a square root to knock controls off of multi-controlled operations until you're left with singly-controlled operations.
Replace controlled-whatevers with controlled-NOTs by finding $A$, $B$, $C$, $\theta$ such that $ABC = I$ but $AXBXC e^{i \theta} = U$ for each single-qubit operation $U$.


The details of these kinds of constructions are covered in textbooks.