What is output amplitude of Direct Digital Sythesizers as AD9850?
In addition to the normal attenuation at increasing frequencies (as @ChrisStratton and OP pointed out), there is the potential to unwittingly add a low-pass filter to the output by decoupling it. This is based on the fact that the trace (or wire) between the DDS output and the decoupling capacitor will have some resistance, which combines with the capacitor to form an RC filter. The cutoff frequency is defined as:
$$
f_{c}=\frac{1}{2 \pi RC}
$$
Taking an arbitrary trace resistance of \$60 m\Omega\$ (consistant with a ½ inch, 8 mil, 0.5 oz trace) with your 100nF capacitor, we get:
$$
f_{c}=\frac{1}{2 \pi \cdot 0.06 \cdot 100n} \approx 26.5MHz
$$
This will change based on the actual resistance, as well as the capacitance of the scope probe (as discussed by @ChrisStratton), but it shows that the capacitance could be having a very real effect on the output amplitude at higher frequencies.