Peak to peak amplitude of sum of sinusoidals (harmonic frequencies)
Suppose that we have a continuous-time periodic signal
$\displaystyle s(t) = a_0 + \sum_{n=1}^N a_n \cos(n\omega_0 t + \theta_n)$.

What does this mean? Do we have a  trace of $s(t)$ on some recorder chart
and the trace looks periodic?  If we did, the question to be solved would be
trivial since we could simply measure the maximum and minimum values of $s(t)$ on
the chart.  So the problem of interest is:


  Given the values of 
  $\omega_0, a_0, a_1, a_2, \cdots, a_N, \theta_1, \theta_2, \cdots, \theta_N$,
  find the values of 
  $$M_\max = \max s(t)\quad \operatorname{and} \quad M_\min =  \min s(t).$$


The OP wants to find the peak-to-peak amplitude of $s(t)$ and this is, of
course, just $M_\max-M_\min$.  It is worth noting that since $s(t)$ is periodic
with period $T = \frac{2\pi}{\omega_0}$, the maximum value and minimum value
occur (at least once) in each interval of length $T$ on the time axis.  The key
issue here is since we know the $a_i$ and the $\theta_i$ etc., we can write
down the formula
$$s(t) = a_0 + \sum_{n=1}^N a_n \cos(n\omega_0 t + \theta_n)\tag{1}$$
(whether this constitutes "reconstruction" of the time-domain signal
is a question that I will leave to others to answer) but the
standard method for finding the maxima and minima of $s(t)$
requires us to find solutions to the nonlinear equation $(2)$ below
(that is, values of $t$ for which $(2)$ holds):
$$\frac{ds(t)}{dt} = -\sum_{n=1}^N a_n\cdot n\omega_0 \sin(n\omega_0 t + \theta_n) = 0.
\tag{2}$$
Once we have found the values $t_1, t_2, \ldots, t_k, \ldots$ for which
$\displaystyle \sum_{n=1}^N a_n\cdot n\omega_0 \sin(n\omega_0 t_k + \theta_n) = 0$,
we can simply calculate $s(t_1), s(t_2), \ldots, s(t_k), \ldots$ and search
through this list of numbers (it is not necessary to sort the list
as recommended in hotpaw2's answer) to find $M_\max$ and $M_\min$.  But,
ignoring the difficulty of finding the $t_k$'s makes hotpaw's answer not
particularly useful in solving the OP's problem.  The answer by ethereal
is even worse in this regard since all it boils down to is the assertion
that (for $\alpha &gt; 0$)
the maximum and minimum values of $\alpha\cdot s(t)$ are $\alpha\cdot M_\max$ 
and $\alpha\cdot M_\min$ respectively. Finding the maximum and minimum
of $s(t)$ from knowledge of its Fourier series is a nontrivial task, not at
all as easy as it is made out to be by ethereal or by hotpaw2.



A related problem has received a lot of research attention in the past
thirty years or so.  For a periodic signal such as $s(t)$, the average power
in the signal is readily computed as 
$$\bar{P} = \frac{1}{T}\int_0^T |s(t)|^2\,\mathrm dt = a_0^2 + \frac{1}{2}\sum_{i=1}^N |a_i|^2.$$
On the other hand, the peak power is $P_\max = \max\{M_\max^2, M_\min^2\}$,
and is also of interest, especially to system and power amplifier designers,
and the ratio $\frac{P_\max}{\bar{P}}$, aptly named the _peak-to-average-power 
ratio (PAPR) has received much attention. It can be calculated exactly for
a given $s(t)$ but only with great computational effort, and so a lot of effort
has gone into finding bounds on the PAPR.  Some of these could help in getting
bounds for the OP's problem.
only with a great deal of effort.