I know that the deviation of difference between the number of heads and tails. With this in mind, I find that the probability that the difference is exactly $k$ increases and then decreases as the number of tosses of the coin increases. Here, I could infer that for an arbitrary interval $[k,l]$, the probability that the difference is included in this interval also increases and then decreases.
However, the expression is too complicated and I do not know how to solve it, so I ask a question.
By symmetry, the average (signed) difference between the number of heads and number of tails must be 0. After all, if you simply interchange heads $\leftrightarrow$ tails, the mathematics is just the same.
So I'm interpreting your question to be:
That answer is:
$$2^{1-n} \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right) \binom{n}{\left\lfloor \frac{n}{2}\right\rfloor +1}$$
Here's a graph:
As expected, this number increases as $n$ increases. After all, if you flip the coin $10^9$ times, you will (on average) get a large absolute difference between the number of heads and the number of tails.
Alternatively, if you want to know that absolute difference as a fraction of $n$, then simply divide the above expression by $n$. Now we see that this expected proportional value approaches zero as $n$ gets large:
Makes perfect sense: At extremely large $n$, the Law of Large Numbers "balances out" the number of heads and tails, so the absolute difference, divided by the total $n$, goes to zero.