As pointed out by E. Schmidt, the sequence A023042 shows that a large percentage of cubes $N^3$ are a sum of three positive cubes. OEIS lists only $N<1770$, but we can extend that:
$$\begin{array}{|c|c|}
\hline
N&\text{%}\\
\hline
2000&85.8\text{%}\\
4000&89.8\text{%}\\
6000&92.1\text{%}\\
8000&93.3\text{%}\\
10000&94.2\text{%}\\
\hline
\end{array}$$
This means that $94.2\text{%}$ of all $N<10000$ have a solution to $a^3+b^3+c^3=N^3$ in positive integers. Note that $N=10000$ is still small. Extrapolating the table, it can be seen that the percentage may easily reach $99\text{%}$ if we go into the millions.
Thus, if we pick a random $N$ in the high end of the range, there is a very good chance that there is an $a,b,c$. For the next perfect number $N=8128$, it is just mere statistics that suggests $N^3$ will be the sum of three positive cubes, and not because it is perfect. In fact, like $496$, it is in several ways,
$$2979^3 + 4005^3 + 7642^3 = 8128^3$$
$$2^6(102^3 + 673^3 + 2007^3) = 8128^3$$
$$2^9(197^3 + 198^3 + 1011^3) = 8128^3$$
And it was almost certain for the next perfect number which is in the millions,
$$2^{27}(3042^3 + 56979^3 + 45845^3) = 33550336^3$$
$$2^{30}(821^3 + 32590^3 + 8227^3) = 33550336^3$$
$$2^{36}(4543^3 + 6860^3 + 5104^3) = 33550336^3$$
Both can be expressed in many more ways than this, and I have only chosen a sample. For the cube of the next perfect number, or $137438691328^3$, chances are even greater that it is a sum of three positive cubes in many ways as well.
Update: Yes, it is:
$$2^{54}(425664^3 + 358719^3 + 275140^3)= 137438691328^3$$
$$2^{54}(432204^3 + 386604^3 + 177535^3)= 137438691328^3$$
Note: Jarek Wroblewski has calculated $a^3+b^3+c^3 = d^3$ with $\color{brown}{\text{co-prime}}$ $a,b,c$, and $d<1000000$ in his website. Using his database and some help with Mathematica and Excel, I came up with the table above which counts all $N$, regardless of whether $a,b,c$ is co-prime or not.
P.S: An interesting question, I believe, is: "Are there infinitely many $N^3$ (especially for prime $N$) that cannot be expressed as a sum of three positive cubes?"
For example, there are no positive integers,
$$a^3+b^3+c^3 = 999959^3$$
even though the percentage of $N<1000000$ with solutions should be close to $99\text{%}$.
In the paper Characterizing the Sum of Two Cubes, Kevin Broughan gives the relevant theorem,
Theorem: Let $N$ be a positive integer. Then the equation $N = x^3 + y^3$ has a solution in positive integers $x,y$ if and only if the following conditions are satisfied:
There exists a divisor $m|N$ with $N^{1/3}\leq m \leq (4N)^{1/3}.$
And $\sqrt{m^2-4\frac{m^2-N/m}{3}}$ is an integer.
The sequence of integers $F(n)$,
$$\begin{aligned}
F(n)
&= a^3+b^3 = (2 n + 6 n^2 + 6 n^3 + n^4)^3 + (n + 3 n^2 + 3 n^3 + 2 n^4)^3\\
&= c^3+d^3 = (1 + 4 n + 6 n^2 + 5 n^3 + 2 n^4)^3 + (-1 - 4 n - 6 n^2 - 2 n^3 + n^4)^3
\end{aligned}$$
for integer $n>3$ apparently is expressible as a sum of two positive integer cubes in exactly and only two ways.
$$\begin{aligned}
F(4) &= 744^3+756^3 = 945^3+15^3\\
F(5) &= 1535^3+1705^3 = 2046^3+204^3\\
&\;\vdots\\
F(60) &=14277720^3+26578860^3 = 27021841^3+12506159^3
\end{aligned}$$
Using Broughan's theorem, I have tested $F(n)$ from $n=4-60$ and, per $n$, it has only two solutions $m$, implying in that range it is a sum of two cubes in only two ways. Can somebody with a faster computer and better code test it for a higher range and see when (if ever) the proposed statement breaks down? Incidentally, we have the nice relations,
$$a+b = 3n(n+1)^3$$
$$c+d = 3n^3(n+1)$$
Note: $F(60)$ is already much beyond the range of taxicab $T_3$ which is the smallest number that is the sum of two positive integer cubes in three ways.
$$T_3 \approx 444.01^3 = 167^3+436^3 = 228^3+423^3 = 255^3+414^3$$
(Using the theorem, this yields 3 values for $m$.)
Best Answer
The lowest four such sums are:
$\begin{align*} 1009=10^3+2^3+1^3&=9^3+6^3+4^3\\ 1366=11^3+3^3+2^3&=9^3+8^3+5^3\\ 1457=11^3+5^3+1^3&=9^3+8^3+6^3\\ 1520=11^3+5^3+4^3&=10^3+8^3+2^3 \end{align*}$
These sums and further examples (though not their decomposition into three distinct positive cubes) are at OEIS. Note, however, that that OEIS sequence includes numbers that are the sum of three distinct positive cubes in different ways where the same cube appears as a term in two ways for the same sum, e.g. $$x^3+1729=x^3+12^3+1^3=x^3+10^3+9^3$$