You are running an old version of Postgres with the setting escape_string_warning = on (default) and standard_conforming_strings = off (outdated!, default is on since Postgres 9.1). The manual:
escape_string_warning(boolean)
When on, a warning is issued if a backslash (\) appears in an ordinary
  string literal ('...' syntax) and standard_conforming_strings is off.
  The default is on. (...)
To just fix the syntax and get rid of the WARNING:
trim(regexp_replace(name, E'\\s\\s+', ' ', 'g'))
Proper solution: Upgrade to a current version of Postgres, or fix the outdated setting to standard_conforming_strings =on.
In modern Postgres, the expression you have is valid as is.
To be precise, \s is the class shorthand for [[:space:]], which includes any kind of white space (incl. tab, nbsp etc.). Your expression replaces any string of two or more white space char with a single space char. The expression to fit your description would be:
trim(regexp_replace(name,'  +', ' ', 'g'))
... which works regardless of version and above settings.
Related: