In pursuit of calculating the fourth power, it might be apt to have a look at the general form of expectation of $p$–th power of sum of iid random variables. 


$\rm [I]$ deduces $\mathbf E\left[S_n^p\right]$ based on a combinatorial reasoning.

Namely, 

_Result $1:$_ For $p\in\mathbb N, $ with iid $X_1, X_2, \ldots, X_n,$ the $p$–th moment of sum $S_n:=\sum_{i=1}^n X_i$ is given by $$\mathbf E\left[S_n^p\right] = \sum_{q_m\in\mathcal Q^p}a_mq_m,\tag 1\label 1$$ where \begin{align}\mathcal Q^p&:=\left\{\underbrace{\mu_{p_1}\mu_{p_2}\cdots\mu_{p_m}}_{:=q_m}\mid p_1, p_2,\ldots, p_m\in \mathbb N_p~\wedge~\sum_{i=1}^m p_i= p\right\},\\a_m&:=\frac{1}{l_1!\cdots l_h!}\frac{n!}{(n-m)!}\frac{p!}{\prod_{i=1}^m p_i!} ,\end{align}

where $h$ is the number of distinct $p_i$s; $l_j$ is the number of times the $j$–th constant appears.

$\rm[II]$ also builds upon multinomial theorem a similar formula, that is equivalent to $\eqref 1.$

_Result $2:$_ $$\mathbf E\left[S_n^p\right] =\sum_{1\leq r\leq p} A_{r,p}{n\choose r},\tag 2\label 2$$ where $$A_{r,p}:= \sum_{\boldsymbol k\in I_{p, r}}{p\choose {k_1\ldots k_r}}{r\choose {n_1\ldots n_a}}\prod_{i=1}^a \mu_{\kappa_i}^{n_i},$$

where $I_{p,r}:= \{\boldsymbol k\in\mathbb N^r\mid k_1+\cdots k_r = p~\wedge~k_1\geq\cdots\geq k_r \}$ is the set of unordered partitioned of $p.$

Deducing the two forms aren't hard; I'm outlining the sketch of the proofs of the corresponding two results above; it is then upto OP to utilize those to formally perform the required calculations: 

_$1:$_ The strategy is to first calculate the cardinality of $\mathcal C_{i_{p_1}, j_{p_2}, \ldots k_{p_m}},$ the set of all unique permutations of the sequence where $X_i$ appears $p_1$ times and so on. This would be $\frac{p!}{\prod_{i=1}^m p_i!} .$ Then we  would calculate the cardinality of unions of all such sets with varying $i, j, \ldots, k,$ no indices being equal to each other. This would yield the factor $\frac{n!}{(n-m)!}.$ Finally we need to take care of the cases where some of $p_i$s can be equal to each other. This propels the factor  $\frac{1}{l_1!\cdots l_h!}.$

$\blacksquare$

$2:$ Crux is expressing the multinomial theorem (with $J_{p, r}$ being the unordered counterpart of $I_{p, r}$ ) as $$\left(\sum_{i=1}^n x_i\right)^p = \sum_{1\leq r\leq p}\sum_{\boldsymbol a \in\mathbb N^r_n}\sum_{\boldsymbol k\in J_{p, r}} {p\choose {k_1\ldots k_r}}\prod_{s=1}^r x_{a_s}^{k_s};$$ intimidating it might look, it is nothing but writing the LHS as  $\sum_{\boldsymbol j \in \mathbb N_{n}^p}\prod_{s=1}^p x_{j_s}$ and breaking the sum based on the distinct elements of $\boldsymbol j$ viz. $\boldsymbol a\in\mathbb N_{n}^r,~r\in \mathbb N_p $ with corresponding multiplicities $\boldsymbol k \in J_{p, r}.$ Now owing to $X_i$s being iid, $\mathbf E\left[X_{a_s}^{k_s}\right] = \mu_{k_s}, ~1\leq s\leq r.$ Thus the final sum isn't impacted by any arbitrary permutation of $k_s$s: take the corresponding $a$ unique parts $\kappa_i$s which appear with multiplicities $n_i$s respectively. This means there are ${r\choose {n_1\ldots n_a}}$ ordered partitions corresponding to a single $\boldsymbol k$ implying $$\sum_{\boldsymbol k\in J_{p, r}} {p\choose {k_1\ldots k_r}}\prod_{s=1}^r \mu_{k_s}=\sum_{\boldsymbol k\in I_{p, r}}{p\choose {k_1\ldots k_r}}{r\choose {n_1\ldots n_a}}\prod_{s=1}^a \mu_{\kappa_s}^{n_s}.$$

$\blacksquare$

$\eqref 1$ seems more tractable. 

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## References: ##

$\rm [I]$ Moments of Sums of Independent and
Identically Distributed Random Variables, _Daniel M. Packwood,_ $2012,$ url: http://arxiv.org/abs/1105.6283.

$\rm [II]$ Moment equalities for sums of random variables via integer partitions and Faàdi
Bruno’s formula, _Dietmar Ferger,_ Turkish Journal of Mathematics: Vol. $38:$ No. $3,$ Article $15,~2014,$ url: https://doi.org/10.3906/mat-1301-6.