Consider the terms of the binomial expansion used to approximate $2.99^5$ and $3.01^5$.
Why do the first three terms of $(3-0.01)^5$ give a better approximation of $2.99^5$ than the first three terms of $(3+0.01)^5$ do for $3.01^5$? I feel like it might have to do with the fact that in the expansion of $(3-0.01)^5$ the signs are alternating between adding and subtracting terms, but I don't know why that leads to a better approximation.
You're right that it's because of the alternating signs. In order to make the question a little less trivial, I'm going to generalise. If $0 < \varepsilon < x$, and $n, k \in \Bbb{N}$ such that $k \le n$, then according to the binomial theorem, $$\left|(x + \varepsilon)^n - \sum_{m=0}^k \binom{n}{m}x^{n-m}\varepsilon^m\right| = \left|\sum_{m=k+1}^n \binom{n}{m}x^{n-m}\varepsilon^m\right| = \sum_{m=k+1}^n \binom{n}{m}x^{n-m}\varepsilon^m.$$ On the other hand $$\left|(x - \varepsilon)^n - \sum_{m=0}^k (-1)^m\binom{n}{m}x^{n-m}\varepsilon^m\right| = \left|\sum_{m=k+1}^n (-1)^m\binom{n}{m}x^{n-m}\varepsilon^m\right| \le \sum_{m=k+1}^n \binom{n}{m}x^{n-m}\varepsilon^m,$$ by the triangle inequality. As such, we can see that the $k$th estimate of $(x - \varepsilon)^n$ is no further than the corresponding estimate of $(x + \varepsilon)^n$ from their respective true values.
This doesn't mean the estimate for $(x - \varepsilon)^n$ is strictly closer. However, in order to achieve equality in the triangle inequality (in $\Bbb{R}$), all terms must have the same sign. So, so long as $k < n - 1$, there will be at least two terms in the sum, each with opposite signs, meaning the inequality will be strict! That is, the estimate for $(x - \varepsilon)^n$ will be strictly closer than $(x + \varepsilon)^n$, excepting the cases $k = n - 1, n$.