If you assume (and it's safe to so) that the metastabillity will settle by the next clock edge - then the output of the second DFF will be a single clock pulse.
I assume that you are discussing asynchronous input signals, slow compared to the clock and without glitches. This can create metastability but probably not the scenario you describe.
A metastable output can not settle at a "wrong value" because there is no wrong value. Since the input changed at the wrong time both output values are correct. Metastability is not settling to the wrong value, it is when the settling is delayed (by a random time). Until it settles, the value seen by following logic is random.
The error happens when a metastable signal goes to more than one logic input and not all inputs agree on the value. If you have inputs connected to the same signal and some see a '0' and other see a '1', you have a problem.
The second synchronizer stage (DFF2) only has one input, so a single clock cycle with a metastable input from DFF1 is normally not a problem.
If the signal sampled by DFF2 is 000X111, the output will be 0000111 or 0001111. Both alternatives are clean, without a single cycle glitch. A correct design should handle both sequences without problems.
The single cycle glitch you describe from DFF2 can only happen if the output from DFF1 is metastable for 2 clock cycles, so that DFF2 can settle at a random value for 2 consecutive clocks. It is not impossible, but unlikely.
The real danger is that the output of DFF2 also goes metastable and doesn't settle in time to meet the setup time of the internal registers in the design.
Note that Altera recommends 3 synchronizer stages if the clock frequency and the data change rate are high.
This will not fix a single cycle glitch created in the first two stages, but it will fix the more realistic problem: that the output of the second stage goes metastable for one clock cycle.