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Use a label expression with python parser. Just replace º with whatever char you want to use:

def FindLabel ([X]):
  s = int([X])
  label = ''.join(['º' for i in range(s)])
  return label

To get a line break every fifth charachter try:

def FindLabel ( [X]  ):
  s = int([X])
  label = ['●' for i in range(s)]
  i = 5
  while i < len(label):
    label.insert(i, "\r\n")
    i += 6
  return ''.join(label)