Let $SX$ be the suspension of CW complex. What are some results available to determine the homotopy groups of $SX$?
This, in general, an incredibly difficult problem. Even we just want to compute the rational homotopy groups of the suspension of $X$ and $X$ is simply connected, where we can do everything using rational homotopy theory to reduce things to commutative DGAs, this starts to involve things like free Lie algebras and the like.
Locally at the prime 2, there is actually a famous long exact sequence when $X$ is a sphere called the EHP long exact sequence. It relates the suspension homomorphism, the Hopf map, and a "whitehead product" map. This gives rise to the EHP spectral sequence that, funnily enough, starts with the 2local homotopy groups of odddimensional spheres and computes the 2local homotopy groups of spheres. Miller and Ravenel have a paper titled "Mark Mahowald's work on the homotopy groups of spheres" that covers some of this material in detail.
Another approach is to say: The "stable" homotopy groups of $X$ are a firstorder approximation using the Freudenthal suspension theorem that Andrea mentioned. There is then a "quadratic" correction term that you can try to use to get an approximation of the homotopy groups that is correct out to roughly three times the connectivity, and so on. These lead into the subject of Goodwillie calculus.
For a classifying space $K(G,1)$, neither of these approaches work very well, because the higher homotopy groups are going to depend pretty intricately on your group itself. For instance, $\pi_3$ of the suspension of the classifying space of a free group is the set of symmetric elements in $G_{\text{ab}}\otimes G_{\text{ab}}$ where $G_{\text{ab}}$ is the abelianization of $G$, and for a general group it lives in an exact sequence between something involving such symmetric elements and the second group homology of $G$. I don't know a closed form for it but maybe someone else knows better.
EDIT: Let me at least be precise, there's an exact sequence $$\pi _4 (\Sigma BG)\rightarrow H_3 G \rightarrow (G_{\text{ab}}\otimes G_{\text{ab}})^{\mathbb Z/2} \rightarrow \pi_3 (\Sigma BG)\rightarrow H_2 G\rightarrow 0$$ Note that for $R$ a ring, an element of $(G_{\text{ab}}\otimes G_{\text{ab}})^{\mathbb Z/2}$ gives rise to an $R$valued symmetric bilinear pairing on $\mathsf{Hom}(G_{\text{ab}},R)$.
EDIT FOR THE FINAL TIME: sorry for the multiple revisions, switching back and forth between homology and cohomology gave me errors. the exact sequence above should be correct now.

$\begingroup$ The functor on abelian groups A > (symmetric elements in A tensor A) is not right exact, and has a 0th right derived functor Γ<sup>2</sup>. This functor is also the "2nd divided square" construction. Is Γ<sup>2</sup> G<sub>ab</sub> the closed form you want? $\endgroup$ – Charles Rezk Oct 28 '09 at 16:36

$\begingroup$ You are right, this "symmetric elements" functor preserves surjections but not coequalizers, but I think the functor I want is its 0'th left (nonabelian) derived functor. $\endgroup$ – Tyler Lawson Oct 29 '09 at 12:27
I think the situation rationally is less hopeless than Tyler Lawson's answer indicates. In fact any simply connected suspension is rationally homotopy equivalent to a wedge of spheres (this is part of Theorem 24.5 in FelixHalperinThomas), and for $\Sigma X$ to be simply connected it suffices that $X$ be connected; in particular this covers the case of a classifying space $X = BG$.
It follows that the dimensions of the rational homotopy groups $\pi_i(\Sigma X, \mathbb{Q})$ are determined by the Betti numbers $b_i = \dim H_i(X, \mathbb{Q})$. For starters, the suspension isomorphism gives $b_i = \dim H_{i+1}(\Sigma X, \mathbb{Q})$ ($i \ge 1$). Hence $\Sigma X$ is rationally homotopy equivalent to a wedge of spheres where $S^{i+1}$ appears $b_i$ times. But the rational homotopy groups of a wedge of spheres is known: as a Lie algebra under the Samelson bracket,
$$\pi_{\bullet}(\Omega \Sigma X, \mathbb{Q}) \cong \pi_{\bullet+1}(\Sigma X, \mathbb{Q})$$
is the free graded Lie algebra on $b_i$ generators of degree $i$, and it's possible to compute its Poincare series. More generally, if $V$ is a graded vector space (which here is $H_{\bullet}(X, \mathbb{Q})$), then by graded PoincareBirkhoffWitt,
$$T(V) \cong U(L(V)) \cong S(L(V))$$
where $T$ denotes the tensor algebra, $L$ denotes the free Lie algebra, $U$ denotes the universal enveloping algebra, and $S$ denotes the symmetric algebra. (Incidentally, $T(V)$ here is $H_{\bullet}(\Omega \Sigma X, \mathbb{Q})$.) The second isomorphism is only an isomorphism of graded vector spaces. Taking Poincare series gives
$$\frac{1}{1  \sum b_k t^k} = \prod \frac{(1 + t^{2k+1})^{\dim L_{2k+1}(V)}}{(1  t^{2k})^{\dim L_{2k}(V)}}.$$
where $\dim L_k(V)$ denotes the dimension of the $k^{th}$ graded component of $L(V)$. These numbers are determined by the $b_i$ a little indirectly as follows. If we write
$$\frac{1}{1  \sum b_k t^k} = \exp \left( \sum_{k \ge 1} \frac{p_k}{k} t^k \right)$$
then
$$\dim L_k(V) = (1)^k \frac{1}{k} \sum_{d  k} (1)^{d/k} \mu(d) p_{d/k}.$$
Example. The most familiar case of this computation to combinatorialists is when the Betti numbers are concentrated in some even degree $b_{2n} = b$ (so that we are computing the rational homotopy of the loop space of a wedge of $b$ copies of $S^{2n+1}$). Then the Poincare series is
$$\frac{1}{1  b t^{2n}} = \exp \left( \sum_{k \ge 1} \frac{b^k}{k} t^{2n k} \right)$$
and we get
$$\dim L_{2n k}(V) = \frac{1}{k} \sum_{d  k} \mu(d) b^{d/k}.$$
These are, up to reindexing, the necklace polynomials.
These identities may look complicated but they give fairly precise asymptotics. For example, if $1  \sum b_k t^k$ is a polynomial and we write it as $\prod (1  r_k t)$, then the $r$ of greatest absolute value occurring in this expression controls the leading term of the asymptotics:
$$\dim L_k(V) \sim \frac{r^k}{k}.$$

$\begingroup$ Thus suspension of 3manifold $M$ is wedge of spheres  I assume it is one $S^4$ and couple of $S^3$ and $S^2$ appears in the wedge in this case. Can we say that number of $S^2$ spheres correspond to first betti number $dim H_1M$ ? Thank you $\endgroup$ – Marek Mitros Nov 2 '18 at 9:56
Since suspension of a $k$sphere is a $k+1$sphere, you have a natural suspension homomorphism $\pi_k(X)\rightarrow \pi_{k+1}(SX)$.
Freudenthal suspension theorem says that if $X$ is $n1$ connected, $n\geq 2$, then the suspension homomorphism is actually an isomorphism for $k<2n1$ and surjective for $k=2n1$.
This is called the stable range. It implies that $\pi_{k+n}(S^n(X))$ eventually stabilizes to a group, called the $k^{\text{th}}$ stable homotopy group of $X$. Here $S^n$ is the $n^{\text{th}}$ iterated suspension.
Outside this range the relation is not so clear, even when you take $X$ to be a sphere (although certainly there is some result unknown to me). Have a look to a table of lowdimensional homotopy groups of spheres to see this wild behaviour.

$\begingroup$ Thanks for your information. If X has good connectness, that will be good by the suspension theorem. But if X is not, eg X is the classifying space, are there some approaches? $\endgroup$ – Skye Oct 23 '09 at 13:45
I can't help but mention a cute application of the suspension isomorphism of (co)homology combined with Pioncare duality to show that for a compact closed manifold $M^n$, if the suspension $\sum M$ is homotopic to a closed orientable manifold, then $M$ is a homology sphere. This actually concerns the homology of the suspension, rather than homotopy groups. I apologize if it doesn't help your question.
In general, for a CW complex $M$, we have the suspension isomorphism $\tilde H_i(M)\cong \tilde H_{i+1}(\sum M)$, the reduced homology. And $\tilde H^i(M)\cong \tilde H^{i+1} (\sum M)$ for reduced cohomology. Now if we have Poincare duality on both sides, we would have $\tilde H^i(M)\cong \tilde H^{i+1}(M)$. passing from reduced and non reduced (co)homology tells us $M$ is a homology sphere.
There is a relevent exercise in "Elements of homology theory" by Viktor Vasilʹevich Prasolov on P45.
Edit: $\tilde H^i(M)\cong \tilde H^{i+1}(M)$ for suitable dimensions. e.g. you can start the induction from $H^{n1}(M)=0$ and then $H^{k1}(M)=H^{k2}(M)$ etc.
Over some assumptions we have for example $\pi_3 SK(G,1) \cong ker(\kappa)$, where $\kappa : G \otimes G \to G'$ is the commutator homomorphism and the group $G \otimes G$ is the nonnabelian tensor square.

1$\begingroup$ What are these assumptions? Or can you give at least examples to make it more concrete? $\endgroup$ – Joonas Ilmavirta Mar 18 '15 at 5:22

1$\begingroup$ This is true for every G; see R. Brown, J.L. Loday, Van Kampen theorems for diagrams of spaces, Topology 26 (1987), 311335. $\endgroup$ – Primoz Mar 18 '15 at 9:13

$\begingroup$ Sorry, my mistake, this is true for every group G, like Primoz just said. The assumptions are needed for it to say that $\pi_3 SX \cong ker(\kappa)$ for a space $X$, where $\kappa : \pi_1 X \otimes \pi_1 X \to (\pi_1 X)'$ is the commutator epimorphism. These are: $X$ is pathconnected and $\pi_2 X = 0$. Then you can see that the above follows. $\endgroup$ – Gustavo Mar 19 '15 at 18:50