Here's a heuristic argument. Let's look at pairs $(a, b)$ and the resulting parity of $a^3 + b^3$ $\pmod 8$

Even cubes, $(2k)^3 \equiv 0$

Odd cubes, $(2k+1)^3 \equiv 2k+1$. So they're evenly distributed between ${1,3,5,7}$

An odd sum of cubes comes from the sum of an even and odd cube, so they're evenly distributed among $1,3,5,7$. (representing $1/2$ of integers).

An even sum of cubes comes from either two evens or two odds. The (even + even) pairs are equivalent to $0$ ($1/8$ of integers). The (odd + odd) pairs are evenly distributed among the remainders $0, 2, 4$, and $6$ (representing $1/2$ of the integers).

Now we can calculate the expected proportions of even and odd taxicab numbers.

1. Odd Taxicabs (Sums $\equiv 1, 3, 5, 7 \pmod 8$). $1/2$ of all pairs land here, distributed evenly across $1/2$ of the integers.$$\text{Odd Taxicabs} \propto \frac{(1/2)^2}{1/2} =\frac{1}{2}$$

2. Even Taxicabs (Sums $\equiv 0 \pmod 8$). This specific remainder makes up $1/8$ of the integers. The pairs that land here are the (even + even) pairs ($1/4$ of total pairs), plus $1/4$ of the (odd + odd) pairs. $$(\text{Even Taxicabs}\equiv 0) \propto \frac{(1/4+(1/4)(1/4))^2}{1/8} = \frac{25}{32}$$

3. Even Taxicabs (Sums $\equiv 2, 4, 6 \pmod 8$). Together, these three remainders make up $3/8$ of the integers. The pairs that land here are the remaining (odd + odd) pairs. $$(\text{Even Taxicabs}\equiv 2, 4, 6) \propto \frac{((3/4)(1/4))^2}{3/8} = \frac{3}{32}$$

So, the expected proportion of Even Taxicabs is:$$\frac{\text{Even}}{\text{Even} + \text{Odd}} = \frac{25/32+3/32}{25/32+3/32 + 16/32} = \frac{7}{11}\approx63.6\%$$