Glypho, 480 bytes
In the "shorthand" format, it's 120 bytes:
1d+d1+d*+ddd++ddd++1+d11+d*d++d11+1+d++d1-dd+++d1<d>+-d++11+d*d*d+<d>d+d+d<d>+d+1+d1-dd+++d1-<d>+d*1+11+1+d+d*d+<d-+>[o]
An example conversion to "true" Glypho (using the translation of the Java interpreter, which differs slightly from that documented in the esolangs.org page) is:
v># # :: < < <v# #*>*> ##:#**#,<,<: : > > * *v>>v # ##,#, + +:++: ++ ##
*<<*,^,^<<#v<<v#v::v< < <,, +,+,+>>+*,,*+*+*,,*>**^v# #,,:^#vv#>+>+ << > >, , ++*>:
:v<v<^#^#v::v>::>v**v # #::>^>+>>:>:>>>*>>##>*^ *# # vv ,::,<<>:++
*vv*v:v:^vv< > > ,,>>:>: << >+>>^ ^ ^^*^+,+,#::#*:*:: :v v , # #<<#<#**#^,^,+##+**
+**+,:,:::>*<^v< v v+^+^*^^*+<<++##+v#v#++<>:< :* **+ + ^ * *<+<+< *vv+<:^^:::
^+*<<***<^+ ++:+:^##^:>:>+::< > >#>># * >,>, :^ ^>>^##<#,<,*^ *<:<
(using Windows line terminators \r\n) where I tried to disguise it as a 2D language for the Programming Language Quiz.
The basic approach is to push onto the stack a 0 followed by the codepoints in reverse order, and then print them with the loop [o]. In order to golf the pushing, I first push 11 and then I can push a new 11 whenever I want with <d>; the final <d-+> replaces that 11 on the bottom of the stack with the desired 0.
I experimented with various values on the bottom of the stack, and 11 is the only one for which my brute-force searcher was able to find expressions for each of the characters which were no more than 11 bytes each. (12 bytes was taking too long).