EDIT: Following a conversation with Ivan Losev, the situation is clearer now. Consider the principal block of $\mathcal{O}$. Recall two facts:

1) the socle of any Verma module is $L_{w_0}$,

2) taking the socle is a left exact functor.

Thus the socle of any object with Verma flag (in particular a tilting module) is isomorphic to a direct sum of copies of $L_{w_0}$.

Now if $T_x$ is an indecomposable tilting module we have

$\dim Hom(L_{w_0}, T_x) = P_{id, xw_0}(1)$

where $P_{y,z}$ is a Kazhdan-Lusztig polynomial. Thus the socle is simple if and only if $P_{id, xw_0} = 1$ which is the case if and only if the Schubert variety $xw_0$ is rationally smooth.

The formula is a consequence of the more general formula (which Peter reminded me of):

$dim Hom(\Delta_x, T_y) = P_{w_0x,w_0y}(1)$

See e.g. "Tilting exercises" or Soergel's papers on tilting modules.

(I deleted the longer version of the answer, because this seems much cleaner than earlier attempts.)