First, I'd like to blow your mind: Increasing sample size does not decrease the variance of an estimate. What you call variance can, for example, stay essentially flat, even as $n$ goes to infinity. But let's come back to that.
We need to define terms. What you're referring to as "variance" is generally called standard error, the standard deviation of the sampling distribution of the statistic. A statistic's sampling distribution is just the distribution of all possible values the estimate could take, over all possible samples of a given size drawn from the population of interest. So, when you ask why variance decreases with sample size, you're really asking why the sampling distribution of a statistic is wider for smaller $n$ and narrower for larger $n$. For the vast majority of practical statistics, your assumption that this will be the case is correct.
The reason this tends to be the case is, as others have said, the Central Limit Theorem. However, what others have neglected to tell you is that there are many limit theorems. Which one applies depends on a) the family of distributions a statistic's sampling distribution belongs to, and b) the asymptotic behavior of the distribution as $n$ goes to infinity.
The vast majority of statistics have a sampling distribution that is approximately and asymptotically normal, so the famous Central Limit Theorem applies. Normal distributions get skinnier as n increases, for reasons others detail. Having a normal sampling distribution gives a statistic a property called efficiency, which just means its observed value gets closer to its expected value with increasing $n$.
That's exactly what we want. We therefore purposely choose statistics with this property when we have the option. It's easy to assume that all statistics have this property when all statistics you see have it, but I guess that's what you'd call selection bias. (It's also called stats being oversimplified for beginners because it's hard enough as it is!)
A particularly interesting counterexample is a certain function of the Hamming distance, computed between a pair of bivariate normal random vectors that have been converted to ranks without ties. That is, suppose you draw $n$ pairs at random from a bivariate normal population with correlation parameter $\rho$, $(X, Y)$. You replace each real number in $X$ with an integer indicating its relative order in $X$ after sorting the vector (first $= 1$, second $= 2$, and so on), and the same for $Y$.
You then count the number of bivariate observations with equal ranks. (So, the count equals $n$ minus the Hamming distance between $X$ and $Y$.) This count's sampling distribution is approximately and asymptotically Poisson with parameter $np$, where $p$ is the probability of obtaining at least one pair of equal ranks (Zolotukhina and Latyshev, 1987).
It is well known that $np = 1$ when $\rho = 0$ (As first shown by Montmort in 1708). Because a Poisson variable's parameter is equal to both its mean and its variance, the standard error of the count only gets closer and closer to $1$ as sample size increases, when $\rho = 0$ (Rae, 1987; Rae and Spencer, 1991). Cool, right? In general, for $\rho \geq 0$, the variance converges to $1/(1 - \rho)$ as $n$ goes to infinity (Zolotukhina and Latyshev, 1987) under bivariate normality. It does NOT decrease for non-negative $\rho$.