I wonder if this perhaps-naive approach can help you. Let's shift by the mean, so consider $X = \sum_{j=1}^n X_j$ where each $X_j = 0.5$ or $-0.5$ independently with one-half probability each. So $\mathbb{E} X^r$ should be the value you're looking for. Now,
\begin{align}
X^r &= \left(X_1 + \dots + X_n\right)^r \\
&= \sum_{\text{$j_1,\dots,j_r$}} ~~ \prod_{s=1}^r X_{j_s}
\end{align}
where the sum is over all vectors $(j_1,\dots,j_r)$ of indices in $\{1,\dots,n\}$.
Now the expectation of the sum is the sum of the expectations, so let's look at the expectation of one term:
\begin{align}
\mathbb{E} \prod_{s=1}^r X_{j_s}
&= \left(0.5^r\right) \mathbb{E} \prod_{s=1}^r \begin{cases} 1 & X_{j_s} > 0 \\ -1 & \text{otherwise} \end{cases} \\
&= \left(0.5^r\right) \begin{cases} 1 & \text{each index appears in $\vec{j}$ an even number of times} \\
0 & \text{otherwise}. \end{cases}
\end{align}
(Why is this true? We split the product into a product over distinct indices $j$ of $X_j^{m(j)}$, where $m(j)$ is the multiplicity of $j$ in the multiset. The indices are independent Bernoullis, so the expectation of the product is the product of the expectations. Each distinct index's expected product is 1 if that index has even multiplicity and zero if odd.)
So if I've made no mistakes, the expectation you're looking for is equal to $0.5^r$ times the number of vectors in $\{1,\dots,n\}^r$ where each $j \in \{1,\dots,n\}$ appears in the vector an even number of times.
Example: For $r=2$, there are $n$ vectors of all-even multiplicity (one for each index, where that index appears twice), so we get $0.5^2*n = n/4$. For $r=4$, there are $n$ vectors containing just $1$ element and $6{n\choose 2}$ vectors containing two elements twice each (because there are six ways to arrange 2 indices into 4 slots, each taking two). So we get $0.5^4 \left(n + 6{n \choose 2}\right) = \frac{1}{16}\frac{2n + 6n^2 - 6n}{2} = \frac{3n^2}{16} - \frac{n}{8}$ (which is also correct -- it's nice to check).
I didn't get to think about how hard this is to estimate in general (I hope someone much more knowledgable in combinatorics can comment).