The slight issue with Will Sawin's answer above for the second part where he suggests Bertrand's postulate is the case where the prime in between $n/2$ and $n$ equals $n-6$, $n-4$ or $n-2$. Therefore it seems better to use the fact that the largest prime gap less than $15\times10^{18}$ is less than $1526$ so for odd $3068 < n < 15\times10^{18}$ there is definitely a prime, $p$, such that $n-1534\leq p \leq n-8$. As $8\leq n - p\leq 1534 < 4\times10^{18}$, $n - p$ is the sum of 2 distinct primes, neither of which could equal $p$ because they are each at most $1534$ and $p>3068-1534=1534$. This implies that all odd numbers $3068<n<15\times10^{18}$ are the sum of 3 distinct primes. As the largest prime gap less than $3100$ is less than $36$, for odd $88<n<3100$ there is a prime, $q$, such that $n-44\leq q\leq n-8$. As $8\leq n-q\leq 44$, $n-q$ is the sum of 2 distinct primes, neither of which could equal $q$ because they are each at most $44$ while $q>88-44=44$. This implies that all odd numbers $88<n<15\times10^{18}$ are the sum of 3 distinct primes. The odd numbers between $19$ and $87$ can easily be checked.
On pg $305$ of Helfgott's proof 'The Ternary Goldbach Problem', it mentions that in (H. A. Helfgott and David J. Platt. Numerical verification of the ternary Goldbach conjecture up to 8.875 · 1030 . Exp. Math., 22(4):406–409, 2013) there is a prime ladder from $3$ to $8.8\times10^{30}$ with consecutive primes on the ladder having difference less than or equal to $4\times10^{18} - 6$. So if $p$ and $q$ are any consecutive primes on the ladder with $p>q$, then $p-q\leq 4\times10^{18} - 6$. Then for any odd n between $8\times10^{18}$ and $10^{27}$, there exist $p,q$ such that $p$ and $q$ are consecutive primes on the prime ladder, $p>q$ and $q + 8\leq n\leq p + 6$, so $8\leq n - q\leq 4\times10^{18}$. Therefore $n - q$ is the sum of 2 distinct primes $r$ and $s$, and so $n$ is the sum of 3 distinct primes (as $q$ cannot be equal to $r$ or $s$ as $r+s=n - q \leq 4\times10^{18}$ so if $n > 8\times10^{18}$ then $q>4\times10^{18}\geq n-q=r+s$. Remember that if $n \leq 8\times10^{18}$ then the argument at the beginning of this shows that $n$ is the sum of 3 distinct primes). I believe that this shows that all odd numbers in the interval $4\times10^{18}$ and $10^{27}$ are the sum of 3 distinct odd numbers (the result pretty much follows from the work done within Helfgott's proof).