What real numbers do algebraic numbers cover? Hardy and Wright mention ( though don't give a proof ) that any finite combination of real quadratic surds is an algebraic number. For example
$\sqrt{11+2\sqrt{7}}$. Are all finite combinations of cube root, fourth root ... $n^{th}$ root also algebraic ? such as $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$.

Yes, you can unpack them into a polynomial that they satisfy.  For your example, we can write $$x=\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}\\x^3=2+3\sqrt[7]{5+3\sqrt{6}}\\\frac 13(x^3-2)=\sqrt[7]{5+3\sqrt{6}}\\\left(\frac 13(x^3-2)\right)^7=5+3\sqrt 6\\\frac 19\left(\left(\frac 13(x^3-2)\right)^7-5\right)^2-6=0$$
Note:  this uses a previous example in the question.  The approach is the same.  Note that the algebraics are closed under the field operations, so the sum of algebraics is again algebraic.
