Update: By a result of Buchta (Zufallspolygone in konvexen Vielecken, Crelle, 1984; available on digizeitschriften.de) there is a general formula for this expected value, it is
$$1 -\frac{8}{3(n+1)} \bigl( \sum_{k=1}^{n+1} \frac{1}{k} (1 - \frac{1}{2^k}) - \frac{1}{(n+1)2^{n+1}} \bigr) $$
yielding (starting with $n=3$): $11/144$, $11/72$, $79/360$, $199/720$, and so on.
The paper contains in fact a more general result, where the problem is solved for any convex $m$-gon; not just the square.
For asymtotics see other answer(s).
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Old version (highly incomplete and wrong guess)
For $n=3$ the expected value is $11/144$ and for $n=4$ it is $11/72$.
This information is taken from a somewhat recent paper (2004) by Johan Philip where the respective distribution functions are studied in detail. I did not see any mention of exact values for other small values of $n$ there (the asymptocic result given already is mentioned though), so they might be unknown.
I misread the problem as that you just wanted the origin in the convex hull. This is a nice problem whose solution deserves to be better known, and this provides an upper bound for the probability that the convex hull contains the unit ball, and the limiting behavior as $r \to \infty$.
J. G. Wendel;
"A Problem in Geometric Probability,"
Mathematica Scandinavica 11 (1962) 109-111.
Zbl 108.31603
http://www.oeis.org/A008949
A special case ($4$ points on the $2$-sphere) was A6 on the 1992 Putnam exam.
You say both $d$-dimensional sphere and $d$-dimensional ball, though I would say the unit $3$-dimensional ball is bounded by a $2$-sphere. I'll assume you mean the ball in $\mathbb{R}^d$.
Consider the kernel of the map $\mathbb{R}^n \to \mathbb{R}^{d}$ of linear combinations of the points. Generically, this kernel has dimension $n-d$. The origin is contained in the convex hull precisely when this kernel intersects the positive orthant, since then some linear combination can be rescaled to have total weight $1$.
By symmetry, all orthants are hit equally often. (Negating a point reflects the kernel, and the action this generates is transitive on the orthants.) A generic $a$-dimensional subspace of $\mathbb{R}^{n}$ hits $2\sum_{i=0}^{a-1} {n-1 \choose i}$ out of $2^{n}$ orthants, so the probability that the convex hull of $n$ random points contains the origin is
$$2^{-n+1}\sum_{i=0}^{n-d-1} {n-1\choose i}. $$
This is an upper bound for the probability that the unit ball is contained in the convex hull.
The origin is contained in the convex hull with probability $1/2$ when the kernel and its orthogonal complement have the same dimension, when $n=2d$.
Best Answer
Imre Bárány has investigated similar questions, including the asymptotics of $p(k,S)$, the probability that $k$ uniformly chosen points from the convex body $S\subset \mathbb{R}^n$ are in convex position (they are extreme points of their convex hull). In general one can give the bounds $$c_1\le k^{2/(n-1)}\sqrt[k]{p(k,S)}\le c_2$$ for large enough $k$ and constants $c_1,c_2$. I don't think closed form formulas are known for all $k$ even for simple convex sets $S$. See here and the papers in the references. See here for the case when $S$ is the unit ball.