Как найти имена столбцов для всех таблиц во всех базах данных в SQL Server

Попробуйте следующее:

select 
    o.name,c.name 
    from sys.columns            c
        inner join sys.objects  o on c.object_id=o.object_id
    order by o.name,c.column_id

С полученными именами столбцов это будет:

select 
     o.name as [Table], c.name as [Column]
     from sys.columns            c
         inner join sys.objects  o on c.object_id=o.object_id
     --where c.name = 'column you want to find'
     order by o.name,c.name

Или для более подробной информации:

SELECT
    s.name as ColumnName
        ,sh.name+'.'+o.name AS ObjectName
        ,o.type_desc AS ObjectType
        ,CASE
             WHEN t.name IN ('char','varchar') THEN t.name+'('+CASE WHEN s.max_length<0 then 'MAX' ELSE CONVERT(varchar(10),s.max_length) END+')'
             WHEN t.name IN ('nvarchar','nchar') THEN t.name+'('+CASE WHEN s.max_length<0 then 'MAX' ELSE CONVERT(varchar(10),s.max_length/2) END+')'
            WHEN t.name IN ('numeric') THEN t.name+'('+CONVERT(varchar(10),s.precision)+','+CONVERT(varchar(10),s.scale)+')'
             ELSE t.name
         END AS DataType

        ,CASE
             WHEN s.is_nullable=1 THEN 'NULL'
            ELSE 'NOT NULL'
        END AS Nullable
        ,CASE
             WHEN ic.column_id IS NULL THEN ''
             ELSE ' identity('+ISNULL(CONVERT(varchar(10),ic.seed_value),'')+','+ISNULL(CONVERT(varchar(10),ic.increment_value),'')+')='+ISNULL(CONVERT(varchar(10),ic.last_value),'null')
         END
        +CASE
             WHEN sc.column_id IS NULL THEN ''
             ELSE ' computed('+ISNULL(sc.definition,'')+')'
         END
        +CASE
             WHEN cc.object_id IS NULL THEN ''
             ELSE ' check('+ISNULL(cc.definition,'')+')'
         END
            AS MiscInfo
    FROM sys.columns                           s
        INNER JOIN sys.types                   t ON s.system_type_id=t.user_type_id and t.is_user_defined=0
        INNER JOIN sys.objects                 o ON s.object_id=o.object_id
        INNER JOIN sys.schemas                sh on o.schema_id=sh.schema_id
        LEFT OUTER JOIN sys.identity_columns  ic ON s.object_id=ic.object_id AND s.column_id=ic.column_id
        LEFT OUTER JOIN sys.computed_columns  sc ON s.object_id=sc.object_id AND s.column_id=sc.column_id
        LEFT OUTER JOIN sys.check_constraints cc ON s.object_id=cc.parent_object_id AND s.column_id=cc.parent_column_id
    ORDER BY sh.name+'.'+o.name,s.column_id

EDIT
Вот базовый пример получения всех столбцов во всех базах данных:

DECLARE @SQL varchar(max)
SET @SQL=''
SELECT @SQL=@SQL+'UNION
select 
'''+d.name+'.''+sh.name+''.''+o.name,c.name,c.column_id
from '+d.name+'.sys.columns            c
    inner join '+d.name+'.sys.objects  o on c.object_id=o.object_id
    INNER JOIN '+d.name+'.sys.schemas  sh on o.schema_id=sh.schema_id
'
FROM sys.databases d
SELECT @SQL=RIGHT(@SQL,LEN(@SQL)-5)+'order by 1,3'
--print @SQL
EXEC (@SQL)

EDIT Версия SQL Server 2000

DECLARE @SQL varchar(8000)
SET @SQL=''
SELECT @SQL=@SQL+'UNION
select 
'''+d.name+'.''+sh.name+''.''+o.name,c.name,c.colid
from '+d.name+'..syscolumns            c
    inner join sysobjects  o on c.id=o.id
    INNER JOIN sysusers  sh on o.uid=sh.uid
'
FROM master.dbo.sysdatabases d
SELECT @SQL=RIGHT(@SQL,LEN(@SQL)-5)+'order by 1,3'
--print @SQL
EXEC (@SQL)

РЕДАКТИРОВАТЬ
На основании некоторых комментариев, вот версия, использующая sp_MSforeachdb :

sp_MSforeachdb 'select 
    ''?'' AS DatabaseName, o.name AS TableName,c.name AS ColumnName
    from sys.columns            c
        inner join ?.sys.objects  o on c.object_id=o.object_id
    --WHERE ''?'' NOT IN (''master'',''msdb'',''tempdb'',''model'')
    order by o.name,c.column_id'