How stable is the top cell of a Lie group?

Perhaps worth noting here:

There is an old open conjecture, the Hirsch conjecture, which says that a stably parallelizable (closed) smooth $n$-manifold $M$ always embeds in $\Bbb R^{[(3/2)n]}$ (approximately). The conjecture has been verified when $M$ is $[n/4]$-connected.

A compact Lie group is parallelizable so, if the conjecture is true, then the top cell will split off after approximately $[n/2]$-suspensions if the Lie group has dimension $n$.

I don't know the minimal number of suspensions required, but for the classical groups $O(n)$, $U(n)$ and $Sp(n)$ the existence of a bound quadratic in $n$ follows from Miller's stable splittings:

• H. Miller. Stable splittings of Stiefel manifolds. Topology 24 (1985) 411-419.

The result is that there is a splitting of $O(n)$, $U(n)$ and $Sp(n)$ as $\bigvee_{k=1}^n\operatorname{Gr}(k,n)^{\operatorname{ad}_k}$, i.e., as a wedge of Thom spaces over the respective Grassmannians. The number of suspensions required to be able to define the splitting map (up to the level $k$ of the decomposition) is the maximum of the embedding dimensions of the Grassmannians involved, cf. Remark 3.9 of Miller's paper. The full splitting exists after $\max\{2\dim_\mathbb{R}\mathbb{K}\cdot k(n-k)\mid 1\leq k\leq n\}$ suspensions where $\mathbb{K}=\mathbb{R},\mathbb{C},\mathbb{H}$ for the three classical series.

As another reference, the following paper gives a discussion of Miller's splitting for $U(n)$ and shows that the top cell splits off after $n^2$ suspensions (see page 462):

• N. Kitchloo. Cohomology splittings of Stiefel manifolds. J. London Math. Soc. 64 (2001), 457-471.

Both papers give explicit constructions of splitting maps and relate the top cell to the adjoint representation of the respective Lie group. Probably this would be true generally, but I don't know of splitting results for the exceptional groups.

Let me just give the naive answer. The homotopy category of $n$-connected $(n+k)$-dimensional CW-complexes is stable for $k<n-1$, by the Freudenthal suspension theorem etc. So the $(\dim G+3)$-fold suspension splits in general, if $G$ is connected the $(\dim G+2)$-fold suspension suffices, and if $G$ is simply connected suspending $\dim G$ times is OK ($\pi_2 G=0$).

This naive bound gives $n^2+2$ for $U(n)$. As Matthias says $n^2$ is sufficient, which is pretty close. For $O(n)$ the bound is $\frac{n^2-n}{2}+3$, which for $n\geq 6$ improves John's estimate. For $SU(n)$ it is $n^2-1$, which is far from optimal as you indicate. The same happens for $Sp(n)$.