#Java 10, 120 113 112 109 bytes
n->{var r="|";for(char c,i=0;n++>0;r=n%3<1?c+r:n%3>1?r+c:r,n/=3)c=(char)(i++<1?49:i<3?51:i<4?57:i+61);return r;}
-3 bytes by using part of the trick of @Arnauld's JavaScript (ES6) answer,
changing i=0 and i++<1?49:i<3?51:i<4?57:i+61 to i=4 and ++i>9?i+55:i>8?57:++i+43.
Order of output: Highest-to-lowest, |-delimiter, lowest-to-highest.
Explanation:
n->{ // Method with integer parameter and String return-type
var r="|"; // Result-String, starting at the delimiter "|"
for(char c=64, // Character
i=4; // Index, starting at 4
;n++>0 // Loop as long as `n` is larger than 0
// Increasing it by 1 with `n++` at the start of every iteration
; // After every iteration:
r= // Change the result to:
n%3<1? // If `n` modulo-3 is 0:
c+r // Prepend the character to the result
:n%3>1? // Else-if `n` modulo-3 is 2:
r+c // Append the character to the result
: // Else:
r, // Leave `r` unchanged
n/=3) // Integer-divide `n` by 3
c=(char)( // Set the character `c` to:
++i>9? // Increase the index `i` by 1 first with `++i`,
// and if it's 10 or larger now:
i+55 // Set `c` to the `i-10`'th (0-indexed) letter of the UC alphabet
:i>8? // Else-if `i` is 9:
57 // Set `c` to '9'
: // Else (it's lower than 9):
++i+43); // Set `c` to either '1' or '3' based on the index `i`
return r;} // Return the result-String