A common mistake in probability is to think that a distribution is uniform because it looks visually flat when all it's values are near zero. This is because we tend to see that $f(x)=0.001 \approx 0.000001=f(y)$ and yet $f(x)/f(y)=0.001/0.000001=1000$, i.e. a small interval around $x$ is 1000 times more likely than a small interval around $y$.
It's definitely not uniform on the entire real line in the limit, as there is no uniform distribution on $(-\infty,\infty)$. It's also not even approximately uniform on $[-2\sigma,2\sigma]$.
You can see the latter from the 68-95-99.7 rule you seem to be familiar with. If it were approximately uniform on $[-2\sigma,2\sigma]$, then the probability of being in $[0,\sigma]$ and $[\sigma,2\sigma]$ should be the same, as the two intervals are the same length. But this is not the case: $P([0,\sigma])\approx 0.68/2= 0.34$, yet $P([\sigma,2\sigma])\approx (0.95-0.68)/2 = 0.135$.
When viewed over the entire real line, this sequence of normal distributions doesn't converge to any probability distribution. There are a few ways to see this. As an example, the cdf of a normal with standard deviation $\sigma$ is $F_\sigma(x) = (1/2)(1+\mbox{erf}(x/\sqrt{2}\sigma)$, and $\lim_{\sigma\rightarrow\infty} F_\sigma(x) = 1/2$ for all $x$, which is not the cdf of any random variable. In fact, it's not a cdf at all.
The reason for this non-convergence boils down to "mass loss" is the limit. The limiting function of the normal distribution has actually "lost" probability (i.e. it has escaped to infinity). This is related to the concept of tightness of measures, which gives necessary conditions for a sequence of random variables to converge to another random variable.