Let SX be the suspension of CW complex? Are there some theorems to determine the homotopy groups of \pi_n(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_{ab} ⊗ G_{ab} where G_{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} (Σ BG) > H_{3} G > (G_{ab} ⊗ G_{ab})^{Z/2} > pi_{3} (Σ BG) > H_{2}G > 0. Note that for R a ring, an element of (G_{ab} ⊗ G_{ab})^{Z/2} gives rise to an Rvalued symmetric bilinear pairing on Hom(G_{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. 


Since suspension of a ksphere is a k+1sphere, you have a natural suspension homomorphism pi_k(X)>pi_{k+1}(SX). Freudenthal suspension theorem says that if X is n1 connected, n at least 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 kth stable homotopy group of X. Here S^n is the nth 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. 


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. 

