A more devious scheme to bring a dam down:
A secret second dam above the main target dam.
You simply install a second dam above the main dam (as high as possible ideally). You reduce the flow inconspicously so that the main dam manning do not see anything unusual, but slowly the reservoir of your second dam is filling.
Now you wait for the perfect time when the second dam reservoir is filled and the main target dam is nearly filled. If the main dam needs only some filling, you can increase the flow from the second dam.
Your attack commences with destroying the second dam which is purposefully built for being brought down (you have some pivot support columns which are disengaged). The water rushes down and gets faster and faster, converting the stored potential energy from a higher point to kinetic energy. Friction and obstacles will slow down the water masses to a point, but it will still be very fast.
When the water enters the main dam, an effect called the water hammer comes in effect. The main dam does not allow the moving water masses to continue running, so the moving water causes a sudden pressure increase. The incoming water not only causes water to slosh over, it literally pushes the dam crest apart. Result: catastrophic failure.
ADDITION: The original question states that we are on the technological level of the early Roman era, so we should neither expect to have a hydro dam like the Hoover nor a reservoir for a city of million people. It will be more like a dam with the height of metres and the reservoir like a big lake.
Still we can compare dynamic with static pressure. The dam need to withstand static pressure, so we can assume we need approximately a pressure with the same order of magnitude to break the dam.
\begin{eqnarray*}
\rho & = & density(kg \, m^{-3}) \\
g & = & gravitational \; acceleration = 9.81 \approx 10 \; m \, s^{-2} \\
h & = & height \; m \\
v & = & velocity \; m \, s^{-1} \\
mean \; static \; pressure & = & \frac{1}{2} \, \rho \; g \; h (The \; dam \; holds \; this \; pressure) \\
dynamic \; pressure & = & \frac{1}{2} \; \rho \; v^2 \Rightarrow v \approx \sqrt{10*h}
\end{eqnarray*}
Moderate flash flood velocity is 2.6 m/s and a very fast flashflood is in the range of 26 m/s. A moderate flashflood will be held by a 0.6m dam, a worst case scenario of 26 m/s would give an impressive height of 70 m. But the flash flood water will merge with the still water in the reservoir, so an inelastic collision will occur and the water slows considerably down. So the final velocity of the water will be the ratio
$$ r = \frac{flash \; flood \; mass}{reservoir \; mass + flash \; flood \; mass}$$ of the flash flood speed (I also neglected friction and energy dissipation by waves).
Result: If the dam is something like 10 m high and the reservoir is big (10-100 times), even the ugliest flashflood will have no pressure effect. Moderate flash floods can be contained even with small dams. On the other hand, if the dam is only a few meters high and the reservoir has not a much bigger capacity (10 times) than the incoming water, an incoming massive flash flood is able to forcibly remove the dam.