Take the logarithm and note that for large $n$ we may replace the sum by an integral:

$$\lim_{n\rightarrow\infty}\frac{1}{n^r}\log K_r(n)=2\log 2+C_r$$
with an $r$-dependent numerical coefficient $C_r$ given by an integral over the $r$-dimensional unit cube:
$$C_r=\int_0^1 dx_1 \cdots\int_0^1 dx_r \log\left[\sum_{p=1}^r \cos^2(\pi x_p/2)\right]$$
$$\qquad=(2/\pi)^r\int_0^1 \frac{dx_1}{\sqrt{1-x_1^2}} \cdots\int_0^1 \frac{dx_r}{\sqrt{1-x_r^2}}\;\log\left[\sum_{p=1}^r x_p^2\right]$$

*Mathematica* tells me $C_2=(4/\pi)G-2\log 2$, which gives the Catalan constant asymptotics of Kasteleyn's formula (who, incidentally, was my colleague here in Leiden).

A numerical integration gives $C_3=0.287095$, $C_4=0.613413$, $C_5=0.856191$, all of which may or may not be expressable in terms of some named constants unknown to *Mathematica*.

For $r\gg 1$ we can argue as follows: consider $r$ i.i.d. random variables $u_1,u_2,\ldots u_r\in(0,1)$, each with distribution
$$P(u)=\frac{1}{\pi\sqrt{u(1-u)}},$$
normalized to unity. The coefficient $C_r$ equals the expectation value
$$C_r=E\left(\log\left[\sum_{p=1}^r u_p\right]\right).$$
Note that $E(u)=1/2$. By concentration of measure we have
$$C_r\rightarrow\log(r/2),\;\;\text{for}\;\;r\rightarrow\infty.$$
I note that $\log(5/2)=0.92$, already not too far from the exact value of $0.86$.

Collecting results, the large-$n$, large-$r$ asymptotics of $K_r(n)$ is
$K_r(n)\mapsto(2r)^{\displaystyle{n^r}}.$ This final answer could have been obtained directly from the definition of $K_r(n)$ as an $n^r$-term product, simply by substituting $1/2$ for each $\cos^2$.