Since you are using sed already, you could just add a second substitution instead of piping into an extra program.

If the backslash appears before an equal sign, as the final two characters on the line:

var=$(sed '/username=/!d; s///; s/\=$/=/' security/boot.properties)

To remove a final backslash, wherever it is:

var=$(sed '/username=/!d; s///; s/(.*)\/1/' security/boot.properties)

To remove all backslashes:

var=$(sed '/username=/!d; s///; s/\//g' security/boot.properties)

Assumptions/understandings:

• OP has stated the entry in security/boot.properties may include multiple characters
• we want to remove only the last character
• the last character is always in the next-to-last position (per OP’s failed coding attempt with rev | cut -c 2-)

For demonstration purposes:

$ cat bp
ignore=this-line
username={AES}BjW32/nnL0m78BJ/ZOEoVzXmEWzVBVZx+8Q=
ignore=this-line

Focusing on the last character residing in the next-to-last position (and assuming we don’t know the last character) we can make use of ksh's builtin substring capabilities:

$ var=$(sed -n 's/username=(.*)/1/p' bp)

$ var="${var:0:${#var}-2}${var: -1}"

$ typeset -p var
var='{AES}BjW32/nnL0m78BJ/ZOEoVzXmEWzVBVZx+8Q='

Where:

• ${#var} – length of string stored in variable var
• keep in mind first character is in position 0 and if there are n characters then the last position is n-1
• ${var:0:${#var}-2} – substring from position 0 to 2nd from last (ie, ignore last 2 characters)
• ${var: -1} – substring starting from last character to end of string (ie, last character)

If we’re not sure where the last character is located in the string we can prepend OP’s current sed script with an additional replacement:

$ var=$(sed -n 's/\([^]*)$/1/; s/username=(.*)/1/p' bp)
$ typeset -p var
var='{AES}BjW32/nnL0m78BJ/ZOEoVzXmEWzVBVZx+8Q='

Where the 1st replacement reads::

• \ – a literal character followed by …
• ([^]*)$ – capture group consisting of 0 or more non- characters up to the end of the string $

Both of these solutions tested on an Ubuntu 22.04 host running ksh 93:

$ ksh --version
  version         sh (AT&T Research) 93u+m/1.0.0-beta.2 2021-12-17

Use this Perl one-liner. Note that the backslash needs to be escaped like so: \:

var=$(  perl -ne 's{(.*)\}{$1} and print;' security/boot.properties )

The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-n : Loop over the input one line at a time, assigning it to $_ by default.

s{PATTERN}{REPLACEMENT} : Replace regex PATTERN with REPLACEMENT.
(.*)\ : 0 or more characters (captured into the variable $1), followed by a literal backslash ().

See also:

• perldoc perlrun: how to execute the Perl interpreter: command line switches
• perldoc perlre: Perl regular expressions (regexes)

This MAY be what you need:

sed -n 's/username=(.*)[]([^]*)/12/p' security/boot.properties

or maybe you’d be better off with awk (nawk or /usr/xpg[46]/bin/awk on Solaris, not the old, broken default /usr/bin/awk):

awk 'match($0,/username=.*\/) { print substr($0,RSTART+9,RLENGTH-8) substr($0,RSTART+RLENGTH+1) }' file

but without knowing what the input looks like that’s obviously just an untested guess and even if the above work for you there may be (actually, probably are) better ways of getting the output you want.