Does central limit theorem hold for general weakly dependent variables? Say I have $X_{ij}$, $j \le i$ with the property that $X_{ij}$ are centered and identically distributed and $E(X_{ij} X_{ij'}) = o(\exp(-i)))$. Then does $\sum_j X_{ij}$ have Gaussian domain of attraction? 
As a related question, if $X_1, X_2, X_3$ are identically distributed and centered and $E(X_i X_j) = c$, what bound can I get for $E(X_1 X_2 X_3)$ in terms of $c$?
 A: Not necessarily. One has to impose more restrictive mixing and moment conditions. A classical book is:
Ibragimov I.A., Linnik Yu.V. Independent and stationary sequences of random variables
There is a long-standing question asked by Ibragimov: is $\phi$-mixing and finiteness of second moment sufficient for CLT to hold for a stationary sequence?
Also, there are various concepts of dependence. For example, if your r.v.'s are associated (i.e. satisfy FKG inequalities) and the covariance decays as you describe, then CLT holds. 
UPD. As for the second part of your question: you cannot estimate higher-order moments in terms of lower-order ones unless the joint distributions have some special structure.
A: Your double subscripts are extraneous.  Let's consider a simpler situation, where we have a single family of random variables $\{X_i\}$.
As Yuri Bakhtin says above, your condition is not sufficient for a CLT to hold.  Here is a simpler situation, however:  suppose that $X_i$ and $X_j$ satisfy finite-range dependence.  That is, there exists a positive integer $R$ such that if $|i-j| \ge R$, then $X_i$ and $X_j$ are independent.  We will prove a law of large numbers for $\{X_i\}$.  If you're interested, you can push it farther to prove a central limit theorem.  Suppose that $X_i$ has mean $\mu$ for each $i$.
Let $S_N = \tfrac{1}{N} \sum_{i=1}^N X_i$ as usual.  Without loss of generality, we may consider indices only divisible by $R$:  $S_{RN} = \tfrac{1}{RN} \sum_{i=1}^{RN} X_i$.  Let $$S_{RN}^{(k)} = \tfrac{1}{N} \sum_{j=0}^{N-1} X_{Rj+k}$$ for $k= 1, \dots, R$, so that $$S_{RN} = \tfrac{1}{R} \left( S_{RN}^{(1)} + \dots + S_{RN}^{(R)} \right).$$Each sum $S_{RN}^{(k)}$ is comprised of independent random variables, so the classical law of large numbers applies and $S_{RN}^{(k)} \to \mu$ both in probability and almost surely.  Consequently, $S_{RN} \to \mu$.  
Obviously, this argument breaks down when $R = \infty$.  In that case, the problem is no longer trivial and you will have to be more cautious with your assumptions.
