Just to be painfully clear, it only seems to make sense to consider the logarithm of returns, i.e. $X=\log (1+\frac r{100})$ for a simple return of $r\%$ in an arbitrary period because this is what sums when returns are temporally aggregated. A basic property of cumulants is that cumulants of all orders are additive under convolution, for which a proof can be found here here.
So if $X_1$, $X_2$, ... $X_n$ are i.i.d., then all the cumulants of $$Y_n = \sum_{i=1}^nX_i$$ scale linearly with $n$, i.e. $$\kappa_k(Y_n)=n\kappa_k(Y_1).$$ However, I suspect that you are normalizing this sum so that the variance (or volatility) remains constant with increasing $n$. So instead let us consider $$Z_n=\frac{Y_n}{\sqrt n}= \frac 1 {\sqrt n} \sum_{i=1}^nX_i.$$ Another basic property of cumulants is that the $k$th cumulant is homogeneous of order $k$ as to scale. Using both properties together we have $$\kappa_k(Z_n)=\left(\frac 1 {\sqrt n}\right)^k\kappa_k(Y_n)=\left(\frac 1 {\sqrt n}\right)^kn\kappa_k(Y_1)=\frac {\kappa_k(Z_1)}{n^{(k-2)/2}}.$$
(Don't forget that $Z_1=Y_1=X_1$.) Now we can show that the statistics scale as you have described: $$\textrm{variance}=\kappa_2(Z_n)=\kappa_2(Z_1)\propto 1;$$
$$\textrm{skewness} =\frac{\kappa_3(Z_n)}{\kappa_2(Z_n)^{3/2}}=\frac{\frac{1}{n^{1/2}}\kappa_3(Z_1)}{\kappa_2(Z_1)^{3/2}}\propto \frac 1{\sqrt n};$$
$$\textrm{ex. kurtosis}=\frac{\kappa_4(Z_n)}{\kappa_2(Z_n)^2}=\frac{\frac{1}{n}\kappa_4(Z_1)}{\kappa_2(Z_1)^{2}}\propto \frac 1 n.$$
There is no reason this cannot be extended to higher orders, although it works out more directly in terms of cumulants than of moments.