Let $Y_i$ be iid with $\mathbb{P}(Y_i=1)=\mathbb{P}(Y_i=-1)=\frac{1}{2}.$ How can I calculate or estimate the expectation.
$$\mathbb{E}[(\sum_{i=1}^nY_i)^4]$$
I know that for $\mathbb{E}[(\sum_{i=1}^nY_i)^2] =\mathbb{E}[(\sum_{i,j}^nY_iY_j)]$ only the diagonal terms survive due to the independence and the zero expectation of the $Y_i$'s. But a similar kind of argument is harder for the 4-th order case since we have terms like $Y_i^3Y_j$ and $Y_i^2Y_j^2$ which will still factor under expectations but they won't all have expectation zero, so you won't be left with only the diagonal terms.
Edit: the estimate has to be quadratic in $n$ since the bound by $n^4$ is trivial.
Let $S_n\sum_{i=1}^nX_i$ and $m_n:= \mathbb E\left[S_n^4\right]$. Using the fact that $$ S_{n+1}^4=S_n^4+4S_n^3X_{n+1}+6S_n^2X_{n+1}^2+4S_nX_{n+1}^3+X_{n+1}^4,$$ we get, after having taken the expectation and use the fact that $X_n^2=X_{n+1}^4=1$, that $$ m_{n+1}=m_n+4\mathbb E\left[S_n^3X_{n+1}\right]+6\mathbb E\left[S_n^2 \right]+4\mathbb E\left[S_nX_{n+1}^3\right]+1. $$ Using independence between $S_n^3$ and $X_{n+1}$, we get $\mathbb E\left[S_n^3X_{n+1}\right]=0$ and using independence between $S_n$ and $X_{n+1}^3$, we get $\mathbb E\left[S_n^3X_{n+1}\right]=0$. By the computations you did for $\mathbb E\left[S_n^2 \right]$, we get $$ m_{n+1}=m_n+6n+1. $$ Since $m_1=1$, one can find the final expression by writing $m_N=\sum_{n=1}^{N-1}\left(m_{n+1}-m_{n}\right)+1$.