Q: What is the percentage of $n$ up to a bound $N$ such that, $$a^3+b^3+c^3 = n^3\tag1$$ has a solution in positive integers?
The sequence A023042 shows a large percentage. I have extended that to $N=10000$,
$$\begin{array}{|c|c|} \hline N&\text{%}\\ \hline 2000&85.8\text{%}\\ 4000&89.8\text{%}\\ 6000&92.1\text{%}\\ 8000&93.3\text{%}\\ 10000&94.2\text{%}\\ \hline \end{array}$$
This means that $94\text{%}$ of all positive integers $N<10000$ has a solution to $(1)$.
How high does the percentage go? From the trends, can one assume that it may reach $97,98,\,\text{or}\;99\text{%}$ if we go up the millions to $N=1000000$?
Based on Jarek Wroblewski's database of co-prime solutions of equation $$a^3+b^3+c^3=n^3\tag{1}$$ (thanks to link from this answer), statistics of numbers $n$ which cannot be written in the form $(1)$, is in the table:
\begin{array}{|l|l|l|} \hline range: 1 - N & cnt(N) & \% \\ \hline 1 - 10 & 8 & 80\% \\ 1 - 100 & 46 & 46\% \\ 1 - 1000 & 191 & 19.10\% \\ 1 - 10000 & 581 & 5.810\% \\ 1 - 100000 & 1192 & 1.192\% \\ 1 - 1000000 & 1867 & 0.186\% \\ \hline \end{array}
Empirically the $cnt(N)$ from this table can be described by rough evaluation: $$ cnt(N) \approx (2\cdot \log_{10}N)^3. $$
Talking about such $k$-digital numbers, we have:
\begin{array}{|l|c|l|l|} \hline k\mbox{-}digital & range & cnt & \% \\ \hline 1\mbox{-}digital & 1 - 9 & 8 & 88.889\% \\ 2\mbox{-}digital & 10 - 99 & 38 & 42.222\% \\ 3\mbox{-}digital & 100 - 999 & 145 & 16.111\% \\ 4\mbox{-}digital & 1000 - 9999 & 390 & 4.3333\% \\ 5\mbox{-}digital & 10000 - 99999 & 611 & 0.6789\% \\ 6\mbox{-}digital & 100000 - 999999 & 675 & 0.0750\% \\ \hline \end{array}
List of $6$-digital such numbers which have form $\overline{9*****}$:
$ 907517, 909911, 913487, 914237, 918463, 922073, 923713, 926183, 929981, 938257, 944701, 946549, 960037, 961853, 968021, 969011, 981947, 984563, 989783, 991961, 992219, 996671, 997727, 998813, 999959. $
Exhaustive list of these numbers which are $<1000000$:
Note: $1007$ of them are prime numbers.
And here is table which shows how many numbers of the list above are divisible by $p,p^2,p^3$ for first unsolvable primes $p$:
\begin{array}{|l|c|c|c|c|} \hline p & \mbox{divisible by } p & ...\; p^2 & ...\;p^3 \\ \hline 2 & \color{blue}{177} & 41 & 22\\ 3 & 31 &-&-\\ 5 & \color{blue}{53} &-&-\\ 7 & \color{blue}{218} & 52 & 11 \\ 11 & 43 &-&-\\ 13 & \color{blue}{300} & 46 & 4 \\ 17 & 47 &-&-\\ 23 & 31 &-&-\\ 31 & 34 &-&-\\ 37 & \color{blue}{82} &-&-\\ 43 & 39 &-&-\\ 47 & 29 &-&-\\ 59 & 24 &1&-\\ 61 & 13 &-&-\\ 73 & 16 &-&-\\ 79 & 48 &4&-\\ \hline \end{array}
There are also $15$ $n$ div by $2^4$. (Hmm, note that factors $7$, $13$ occur often enough).