From @chepner's comment on my other answer:

position = 0  # or wherever you left off last time
try:
    with open('myfile.txt') as file:
        file.seek(position)  # zero in base case
        for line in file:
            position = file.tell() # current seek position in file
            # process the line
except:
    print 'exception occurred at position {}'.format(position)
    raise

Iterating over a file object yields lines with full line endings intact. You should be able to just add the lens to a counter object to get the position. You'll need to multiply based on character encoding (character byte size)

position = 0  # or wherever you left off last time
try:
    with open('myfile.txt') as file:  # don't you go correcting me on naming it file. we don't call file directly anyway!
        file.seek(position)  # zero in base case
        for line in file:
            position += len(line)
            # process the line
except:
    # yes, a naked exception. TWO faux pas in one answer?!?
    print 'exception occurred at position {}'.format(position)
    raise # re-raise to see traceback or what have you

Well, if your pattern is simple, this would be simple

$ echo -e '#!/bin/bash\necho abracadabra' >/tmp/script
$ pattern=bash
$ sed -rn "0,/$pattern/ {s/^(.*)$pattern.*$/\1/p ;t exit; p; :exit }" /tmp/script \
    | wc -c 
8

As you can see, this will output position of the first character in your pattern, assuming first character in the file has number 1.

NB 1: sed has a habit to add a trailing newline to the last string it parses, thus, when we take a part of the line preceding the pattern, the number of bytes in output should be 7 (count them → #!/bin/), but what wc -c actually counts looks like

$ sed -rn "0,/$pattern/ {s/^(.*)$pattern.*$/\1/p ;t exit; p; :exit }" /tmp/script \
   | hexdump -C
00000000  23 21 2f 62 69 6e 2f 0a                           |#!/bin/.|
00000008

This could be potential source of troubles, if you were looking for EOF, for example. I can’t think of a more appropriate case, I just want to point that out.

NB 2: If pattern will contain special characters, sed will fail. If you could provide an example of what you are looking for, I could escape it.

NB 3: This assumes the pattern is unique. If you will stop reading the file on a second or third instance of pattern, this will not work.

Update. I’ve found a more simple way.

$ grep -bo bash <<< '#!/bin/bash'
7:bash

For GNU grep there are two options:

-b, --byte-offset
    Print the 0-based byte offset within the input file before  each  line  of
    output. If -o (--only-matching)  is specified, print the offset of the
    matching part itself.

I’d suggest to use grep, because if you specify -F key, it will treat pattern as a simple string.

$ grep -F '!@##$@#%%^%&*%^&*(^)((**%%^@#' <<<'!@##$@#%%^%&*%^&*(^)((**%%^@#' 
!@##$@#%%^%&*%^&*(^)((**%%^@#