Probability for the length of the longest run in $n$ Bernoulli trials
This problem was solved using generating functions by de Moivre in 1738. The formula you want is $$\mathbb{P}(\ell_n \geq m)=\sum_{j=1}^{\lfloor n/m\rfloor} (-1)^{j+1}\left(p+\left({n-jm+1\over j}\right)(1-p)\right){n-jm\choose j-1}p^{jm}(1-p)^{j-1}.$$
References
Section 14.1 Problems and Snapshots from the World of Probability by Blom, Holst, and Sandell
Chapter V, Section 3 Introduction to Mathematical Probability by Uspensky
Section 22.6 A History of Probability and Statistics and Their Applications before 1750 by Hald gives solutions by de Moivre (1738), Simpson (1740), Laplace (1812), and Todhunter (1865)
Added: The combinatorial class of all coin toss sequences without a run of $ m $ heads in a row is $$\sum_{k\geq 0}(\mbox{seq}_{< m }(H)\,T)^k \,\mbox{seq}_{< m }(H), $$ with corresponding counting generating function $$H(h,t)={\sum_{0\leq j< m }h^j\over 1-(\sum_{0\leq j< m }h^j)t}={1-h^ m \over 1-h-(1-h^ m )t}.$$ We introduce probability by replacing $h$ with $ps$ and $t$ by $qs$, where $q=1-p$: $$G(s)={1-p^ m s^ m \over1-s+p^ m s^{ m +1}q}.$$ The coefficient of $s^n$ in $G(s)$ is $\mathbb{P}(\ell_n<m).$
The function $1/(1-s(1-p^ m s^ m q ))$ can be rewritten as \begin{eqnarray*} \sum_{k\geq 0}s^k(1-p^ m s^ m q )^k &=&\sum_{k\geq 0}\sum_{j\geq 0} {k\choose j} (-p^ m q)^js^{k+j m }\\ %&=&\sum_{j\geq 0}\sum_{k\geq 0} {k\choose j} (-p^ m q )^js^{k+j m }. \end{eqnarray*} The coefficient of $s^n$ in this function is $c(n)=\sum_{j\geq 0}{n-j m \choose j}(-p^ m q)^j$. Therefore the coefficient of $s^n$ in $G(s)$ is $c(n)-p^ m c(n- m ).$ Finally, \begin{eqnarray*} \mathbb{P}(\ell_n\geq m)&=&1-\mathbb{P}(\ell_n<m)\\[8pt] &=&p^ m c(n- m )+1-c(n)\\[8pt] &=&p^ m \sum_{j\geq 0}(-1)^j{n-(j+1) m \choose j}(p^ m q)^j+\sum_{j\geq 1}(-1)^{j+1}{n-j m \choose j}(p^ m q)^j\\[8pt] &=&p^ m \sum_{j\geq 1}(-1)^{j-1}{n-j m \choose j-1}(p^m q)^{j-1}+\sum_{j\geq 1}(-1)^{j+1}{n-j m \choose j}(p^mq )^j\\[8pt] &=&\sum_{j\geq 1}(-1)^{j+1} \left[{n-j m \choose j-1}+{n-j m \choose j}q\right]p^{ jm } q^{j-1}\\[8pt] &=&\sum_{j\geq 1}(-1)^{j+1} \left[{n-j m \choose j-1}p+{n-j m \choose j-1}q+{n-j m \choose j}q\right]p^{ jm } q^{j-1}\\[8pt] &=&\sum_{j\geq 1}(-1)^{j+1} \left[{n-j m \choose j-1}p+{n-j m +1\choose j}q \right]p^{ jm} q^{j-1}\\[8pt] &=&\sum_{j\geq 1}(-1)^{j+1} \left[p+{n-j m +1\over j}\, q\right] {n-j m \choose j-1}\,p^{ jm} q^{j-1}. \end{eqnarray*}
Define a Markov chain with states $0, 1, \ldots m$ so that with probability $1$ the chain moves from $m$ to $m$ and for $i<m$ with probability $p$ the chain moves from $i$ to $i+1$ and with probability $1-p$ the chain moves from $i$ to $0$. If you look at the $n$th power of the transition matrix for this chain you can read off the probability that in $n$ flips you have a sequence of at least $m$ consecutive heads.
You can find a limiting distribution, otherwise it's a difficult problem and the closed form solution won't have much practical value. See this for an elementary approach. [Update] Previous link moved to this new address. "Longest Run of Heads", M.F.Schilling.