Duplicating Excel-workbook, and transferring data
Since you use continue in the if clause, you can get rid of the else part and save a level of indentation.
You can use a list-comprehension to build the list of r_value: [worksheet.cell_value(row + 1, 1) for row in range(3)].
You can iterate directly over the content of r_value to set it back:

for row, value in enumerate(r_value, 3):
    s.write(row, 1, value)

I don't think you need to copy the template first and modify it after. You may be able to open the template and then save it using the new filename.
You can use a second parameter with a default value to hold the name of the template.


Proposed improvements:

import os

from xlrd import open_workbook
from xlutils.copy import copy


def duplicate_and_transfer(path, template_name='Template.xlsx'):
    template = os.path.join(path, template_name)

    for filename in os.listdir(path):
        if filename == template_name:
            continue

        filepath = os.path.join(path, filename)
        excel_file = open_workbook(filepath)
        worksheet = excel_file.sheet_by_index(0)
        new_filename = worksheet.cell_value(5,2)
        new_filepath = os.path.join(path, new_filename + '.xls')
        values = [worksheet.cell_value(row + 1, 1) for row in range(3)]

        rb = open_workbook(template)
        workbook = copy(rb)
        worksheet = workbook.get_sheet(0)
        worksheet.write(7, 2, new_filename)
        for row, value in enumerate(values, 3):
            worksheet.write(row, 1, value)

        workbook.save(new_filepath)


if __name__ == '__main__':
    duplicate_and_transfer(r'C:\Users\Stewie\Documents\excel_folder')


You may as well be able to open the template once and save it several times.

Lastly, if you have troubles working with XLSX files, you may want to try openpyxl instead:

import os

from openpyxl import load_workbook


def duplicate_and_transfer(path, template_name='Template.xlsx'):
    template = load_workbook(os.path.join(path, template_name))

    for filename in os.listdir(path):
        if filename == template_name:
            continue

        workbook = load_workbook(os.path.join(path, filename))
        worksheet = workbook.worksheets[0]

        new_filename = worksheet['C6'].value
        template['B8'] = new_filename
        new_filepath = os.path.join(path, new_filename + '.xlsx')

        # Using double unpacking as the ranges will return a 3-tuple of 1-tuple
        for (src,), (dest,) in zip(worksheet['B2:B4'], template['B4:B6']):
            dest.value = src.value

        template.save(new_filepath)


if __name__ == '__main__':
    duplicate_and_transfer(r'C:\Users\Stewie\Documents\excel_folder')