This uses quite a few SQL tricks (SQL Server 2005):
CREATE TABLE [dbo].[stackoverflow_165571](
[visit] [datetime] NOT NULL
) ON [PRIMARY]
GO
;WITH buckets AS (
SELECT dateadd(mi, (1 + datediff(mi, 0, visit - 1 - dateadd(dd, 0, datediff(dd, 0, visit))) / 5) * 5, 0) AS visit_bucket
,COUNT(*) AS visit_count
FROM stackoverflow_165571
GROUP BY dateadd(mi, (1 + datediff(mi, 0, visit - 1 - dateadd(dd, 0, datediff(dd, 0, visit))) / 5) * 5, 0)
)
SELECT LEFT(CONVERT(varchar, l.visit_bucket, 8), 5) + ' - ' + CONVERT(varchar, SUM(r.visit_count))
FROM buckets l
LEFT JOIN buckets r
ON r.visit_bucket <= l.visit_bucket
GROUP BY l.visit_bucket
ORDER BY l.visit_bucket
Note that it puts all the times on the same day, and assumes they are in a datetime column. The only thing it doesn't do as your example does is strip the leading zeroes from the time representation.