I will assume that we can use the same digit more than once,
since it is not stated that we cannot.  

First we calculate it as though the 1st digit could be 0.
[We know it can't be 0 for such numbers would only have 3 or
fewer digits.  Hoever we'll subtract those afterwards.]

If the first digit could be 0, there would be 5 ways to
choose each of the 4 digits or 5^4 or 625 "5-digit" numbers,
some with 0's coming first.

This would be a list of 625 numbers to add like this:

0000
0001
...
5555
----
SUM

In any column of that addition there are exactly as many of any
one of the 5 digits as any other.  So there are 625/5 or 125
of each digit in each column.  So the sum of each of the 4
columns is 125(0+1+2+3+5) = 1375.  So the sum of the list above
is 1375∙1000+1375∙100+1375∙10+1375 = 1375(1000+100+10+1) = 1375∙1111 =

1527626.

But from that we must subtract the sum of the "4-digit numbers" which
start with 0 and are actually have 3 or fewer digits.  That is we must 
subtract this sum:

000
001 
...
555
---
SUM

That is a list of 5^3 or 125 numbers to add.

We use the same reasoning as above:

In any column of that addition there are exactly as many of any
one of the 5 digits as any other.  So there are 125/5 or 25
of each digit in each column.  So the sum of each of the 3
columns is 25(0+1+2+3+5) = 275.  So the sum of the list above
is 275∙100+275∙10+275 = 275(100+10+1) = 275∙111 = 30525

Then the final answer is 1527626 - 30525 = 1497100.

[Note, if each of the digits can only be used once, tell me in the
thank you note form below and I'll get back to you by email. No
charge ever. I do this for fun.]

Edwin