This is exactly the problem of finding the expected maximum of $n$ iid geometric (1/2) random variables.  This is because the geometric distribution models the time until the first success in independent Bernoulli trials, and each coin flip can be considered an independent Bernoulli trial.  

The question of finding the expected maximum of $n$ iid geometric random variables [was asked][1] a few months ago on MO.  You can see from the accepted answer there that the expected value can be expressed as

$$\sum_{i=1}^n \binom{n}{i} (-1)^{i+1} \frac{1}{1-\frac{1}{2^i}},$$  
which is the last expression in Didier Piau's answer in a slightly different form.

The other answer cites a paper by Bennett Eisenberg that claims that "There is no... simple expression for... the expected value of the maximum of $n$ IID geometric random variables."  However, the paper itself proves that $E[T(n)] - \sum_{k=1}^n \frac{1}{\lambda k}$ "is very close to 1/2 not only for moderate values of $\lambda$, but also for relatively small values of $n$ and that this difference is logarithmically summable to 1/2 for all values of $\lambda$."  In your case, $\lambda = \ln 2$.  Thus $E[T(n)]$ is very close to $\frac{H_n}{\ln 2} + \frac{1}{2}$, right in the middle of the range given by Louigi Addario-Berry's answer.


  [1]: https://mathoverflow.net/questions/41604/what-is-the-expected-maximum-out-of-a-sample-size-n-from-a-geometric-distributi