This is the mean absolute deviation (MAD) for a binomial distribution, divided by $n$. The expectation is hence $$2 \, (1-p)^{n+1-\lceil np \rceil} \, p^{\lceil np \rceil} \, \binom{n-1}{\lceil np \rceil-1} \;.$$ See this paper for bounds and a proof of the above expression.

This is the mean absolute deviation (MAD) for a binomial distribution, divided by $n$. The expectation is hence $$2 \, (1-p)^{n+1-\lceil np \rceil} \, p^{\lceil np \rceil} \, \binom{n-1}{\lceil np \rceil-1} \;.$$ See this paper (Berend & Kontorovich 2013, doi: 10.1016/j.spl.2013.01.023) for bounds and a reference for the above expression.

Let $r = \lceil n p \rceil$. From experimentation, it seems like the expectation is $$2 \, (1-p)^{n+1-r} \, p^r \, \binom{n-1}{r-1}$$ although I cannot show this..

Edit: This is actually just the mean absolute deviation. See this paper for bounds and a proof of the above expression.

This is the mean absolute deviation (MAD) for a binomial distribution, divided by $n$. The expectation is hence $$2 \, (1-p)^{n+1-\lceil np \rceil} \, p^{\lceil np \rceil} \, \binom{n-1}{\lceil np \rceil-1} \;.$$ See this paper for bounds and a proof of the above expression.