As stated by luchonacho, exhaustion will show that it's not possible in less than 7 digits with those operations.
Not showing negative values, as they are symmetric to positive ones, here are the easy attainable integer values:
1 digit: 1
2 digits: 0, 2, 11, 11!, ...
3 digits: 3, 6, 10, 12, 111, 720, 10!, ...
4 digits: 4, 5, 7, 9, 13, 22, 24, 110, 112, 120, 121, 719, 721, 1111, ...
5 digits: 8, 14, 17, 20, 21, 23, 25, 33, 36, 64, 66, 100, 109, 113, 119, 122, 132, 144, 222, 360, ...
And now we know we're stuck, because to achieve 29 (a prime number) in 6 digits, we would actually need any of those values in 5 digits: 28, 30, 784, 900, ... and we don't have them.
As such, any solution for 29 in 7 digits is optimal:
29 = (11−1)×(1+1+1)−1 (found by Dr Xorile)
Note that there may be room for a non-trivial sum of factorial numbers to reach a square power of 29 (or of 29±1), as found by Amorydai, but it's unlikely to beat 7 digits.
For illustration, while all values below 27 are attainable trivially with 6 digits or less, here are some solutions in 7 digits for numbers greater or equal to 28:
28 in 7 digits = (1+1+1)^(1+1+1)+1
28 in 7 digits = (1+1)×(11+1+1+1)
29 in 7 digits = (11−1)×(1+1+1)−1
30 in 6 digits = (11−1)×(1+1+1)
31 in 7 digits = (11−1)×(1+1)+11
31 in 7 digits = (11−1)×(1+1+1)+1
31 in 7 digits = (1+1+1)×11-1-1
32 in 6 digits = (1+1+1)×11-1
33 in 5 digits = (1+1+1)×11
34 in 6 digits = (1+1+1)×11+1
35 in 6 digits = (1+1+1)!^(1+1)-1
36 in 5 digits = (1+1+1)!^(1+1)
37 in 6 digits = (1+1+1)!^(1+1)+1
38 in 7 digits = (1+1+1)!^(1+1)+1+1
39 in 7 digits = (11+1+1)×(1+1+1)
40 in 7 digits = (11-1)×(1+1+1+1)
40 in 7 digits = (11-1)×(1+1)×(1+1)
15 is the lowest positive value requiring 6 digits with this reasoning.
28 is the lowest positive value requiring 7 digits with this reasoning.
41 is the lowest positive value requiring 8 digits with this reasoning.