Expected flips for $n$ coins 
In the start, $n$ coins are flipped. Every round, if a coin lands
  heads, it stops being flipped. What is the expected number of rounds it will
  take to stop flipping completely?

For $1$ coin, the answer is $2$. For $n$ coins, this is $E(1) + 2E(2) + ...$, with each $E(X)$ meaning one coin took $X$ tries and all the others took $\le X$ tries ($\max = X$). I am unsure of how to find a formula for this.
 A: The probability that a given coin hasn't stopped after $k$ rounds is $(1/2)^k$, so the probability that it's stopped is $1-(1/2)^k$, so the probability that they've all stopped is $\left(1-(1/2)^k\right)^n$, so the probability that they haven't all stopped is $1-\left(1-(1/2)^k\right)^n$. Then the expected number of rounds is
\begin{eqnarray}
\sum_{k=0}^\infty\left(1-\left(1-\left(\frac12\right)^k\right)^n\right)&=&-\sum_{k=0}^\infty\sum_{j=1}^n\binom nj(-1)^j\left(\frac12\right)^{jk}\\
&=&
-\sum_{j=1}^n\binom nj(-1)^j\sum_{k=0}^\infty\left(\frac12\right)^{jk}\\
&=&
-\sum_{j=1}^n\binom nj\frac{(-1)^j}{1-\left(\frac12\right)^j}\;.
\end{eqnarray}
The first few values are $2$ for $n=1$, then $2\cdot2-4/3=2\frac23$ for $n=2$ and $3\cdot2-3\cdot4/3+8/7=3\frac17$ for $n=3$.
A: Let $X_i$ be the number of times we need to flip the $i$-th coin and $X$ the total number of rounds. 
Then, $X = \displaystyle\max_{1 \le i \le n} X_i$. So, we have:
$\Pr[X \ge m]$ $= 1-\Pr[X < m]$ 
$= 1-\Pr[X_i < m \ \text{for all} \ i]$ 
$= 1-\Pr[X_i < m]^n$ 
$= 1-(1-\Pr[X_i \ge m])^n$ 
$= 1-(1-\Pr[m-1 \ \text{heads in a row}])^n$
$= ?$ (I'll let you finish this)
Finally, use the formula $E[X] = \displaystyle\sum_{m = 1}^{\infty}m\Pr[X = m] = \sum_{m = 1}^{\infty}\Pr[X \ge m]$.
EDIT: The values of $E[X]$ for $n = 1,2,3,4,5,6$ are $2, \dfrac{8}{3}, \dfrac{22}{7}, \dfrac{368}{105}, \dfrac{2470}{651}, \dfrac{7880}{1953}$. 
The sequence of numerators and denominators are OEIS A158466 and OEIS A158467 respectively.
A: This is the same result gotten by joriki. I have included a bit more detail of the computation (mainly because it is easier for me to follow). I have also added the asymptotic behavior of the expected duration.

Expected Duration
The probability that after $k$ flips, a particular coin has come up heads, at least once, is $1-\frac1{2^k}$. Thus, the probability that $n$ coins have all come up heads, at least once, is $\left(1-\frac1{2^k}\right)^n$. Therefore, the probability that all coins have come up heads at least once on flip $k$ is $\left(1-\frac1{2^k}\right)^n-\left(1-\frac1{2^{k-1}}\right)^n$. Thus, the expected duration until all coins have come up heads at least once is
$$
\begin{align}
&\sum_{k=1}^\infty k\left[\left(1-\frac1{2^k}\right)^n-\left(1-\frac1{2^{k-1}}\right)^n\right]\\
&=\lim_{N\to\infty}\left[\sum_{k=1}^Nk\left(1-\frac1{2^k}\right)^n-\sum_{k=0}^{N-1}(k+1)\left(1-\frac1{2^k}\right)^n\right]\\
&=\lim_{N\to\infty}\left[N-\sum_{k=0}^{N-1}\left(1-\frac1{2^k}\right)^n\right]\\
&=\sum_{k=0}^\infty\left[1-\left(1-\frac1{2^k}\right)^n\right]\\
&=\sum_{k=0}^\infty\sum_{j=1}^n(-1)^{j-1}\binom{n}{j}2^{-jk}\\
&=\bbox[5px,border:2px solid #C0A000]{\sum_{j=1}^n(-1)^{j-1}\binom{n}{j}\frac1{1-2^{-j}}}
\end{align}
$$

Asymptotic Behavior
As shown above, the expected duration is
$$
\begin{align}
&\sum_{k=0}^\infty\left[1-\left(1-\frac1{2^k}\right)^n\right]\\
&=1+\sum_{k=1}^{\lfloor\log_2(n)\rfloor}\left[1-e^{-n/2^k}+O\left(\frac1{2^k}\right)\right]
+\sum_{k=\lfloor\log_2(n)\rfloor+1}^\infty O\left(\frac{n}{2^k}\right)\\
&=\bbox[5px,border:2px solid #C0A000]{\log_2(n)+O(1)}
\end{align}
$$

To clarify the estimates above, we have the following.
For $a\gt b$, we have $a^n-b^n\le na^{n-1}(a-b)$, therefore,
$$
\begin{align}
e^{-n/2^k}-\left(1-\frac1{2^k}\right)^n
&\le ne^{-(n-1)/2^k}\left(e^{-1/2^k}-1+\frac1{2^k}\right)\\
&=2^k\left(\frac{n}{2^k}e^{-n/2^k}\right)e^{1/2^k}O\left(\frac1{4^k}\right)\\
&=O\left(\frac1{2^k}\right)
\end{align}
$$
Furthermore, since the greatest exponent is $-1$ and they occur with a ratio of $2$,
$$
\begin{align}
\sum_{k=1}^{\lfloor\log_2(n)\rfloor}e^{-n/2^k}
&\le\sum_{k=0}^\infty e^{-2^k}\\[6pt]
&\doteq0.521866
\end{align}
$$
Similarly, since $\frac{n}{2^k}\le1$ and the terms occur with a ratio of $2$,
$$
\begin{align}
\sum_{k=\lfloor\log_2(n)\rfloor+1}^\infty\frac{n}{2^k}
&\le\sum_{k=0}^\infty2^{-k}\\[6pt]
&=2
\end{align}
$$
A: Let $E(j)$ denote the expected number of additional throws when there are $j$ coins still "alive". It is pretty obvious that the $E(j)$ satisfy 
$$E(0)=0,\qquad E(j)=1+\sum_{k=0}^j{j\choose k}2^{-j}\>E(k)\quad(j\geq1)\ .$$
This leads to the recursion
$$E(j)={1\over 2^j-1}\left(2^j+\sum_{k=0}^{j-1}{j\choose k}E(k)\right)\qquad(j\geq1)$$
and to the values found by @joriki and @JimmyK4542.
