Why is the size of Al3+ less than that of Li+?
Now, it seems obvious that the size of Al3+ should be greater than that of Li+


No it does not. It is true that electrons repulse each other, however, they all are attracted towards the nucleus.  The real 'size' of an atom or ion is a result of compromise between electron-electron repulsion and electron-nuclear attraction.


  I think that it is a kind of exception


Obviously, it is not.


  Why is this? 


Commonly, there are three consideration, used to rationalize basic trends in atomic/ionic radii.  


electron-electron repulsion (negative charges repulse each other)
electron-nucleus attraction (negative charges are attracted towards positive ones)
electron shielding (electrons in shells closer to nucleus 'shield' outer electrons from some charge of the nucleus.


While electrons repulse each other, increasing nuclear charge more then compensates it; so if not for shell organisation of electrons, atoms would monotonically decrease in size toward the end of the periodic table, as it is observable in each period 1—3.

However, inner electronic shells effectively shield the outer electrons from the nucleus, so they 'see' a nucleus that is a) larger in size and b) has charge equal to nuclear charge minus number of electrons in inner shells. Given that, outer electrons 'see' a nucleus with low charge density, so outer shells progressively increase in size within the same column. 

In addition to that, electrons still repulse each other, so positive ions are smaller than neutral atoms, and neutral atoms are smaller than negative ions.

Now, consider $\ce{Li+}$ and $\ce{Al^{3+}}$. For the former, 2 outer electrons feel the charge of three protons. For the latter, 8 outer electrons feel the charge of 11 protons in the nucleus. Such great charge easily beats all other factors, so the latter cation is much smaller than the former.