Let's see what QFT does on two qubit (and then on three qubit) computational basis states and try to gain some insights. The QFT action on $|j\rangle$ basis state:
$$QFT |j\rangle = \frac{1}{2^{\frac{n}{2}}} \sum_{k=0}^{2^n -1} e^{2 \pi i \frac{jk}{2^n}} |k\rangle$$
where $n$ is the qubit number. Now suppose $n=2$, then:
\begin{align*}
QFT |00\rangle &= QFT |0\rangle = \frac{1}{2} \sum_{k=0}^{3} e^{2 \pi i \frac{0 \cdot k}{4}} |k\rangle = \frac{1}{2}\big( |0\rangle + |1\rangle + |2\rangle + |3\rangle \big)
\\
QFT |01\rangle &= QFT |1\rangle = \frac{1}{2} \sum_{k=0}^{3} e^{2 \pi i \frac{1 \cdot k}{4}} |k\rangle = \frac{1}{2}\big( |0\rangle + i |1\rangle - |2\rangle - i|3\rangle \big)
\\
QFT |10\rangle &= QFT |2\rangle = \frac{1}{2} \sum_{k=0}^{3} e^{2 \pi i \frac{2 \cdot k}{4}} |k\rangle = \frac{1}{2}\big( |0\rangle - |1\rangle + |2\rangle - |3\rangle \big)
\\
QFT |11\rangle &= QFT |3\rangle = \frac{1}{2} \sum_{k=0}^{3} e^{2 \pi i \frac{3 \cdot k}{4}} |k\rangle = \frac{1}{2}\big( |0\rangle - i|1\rangle - |2\rangle + i|3\rangle \big)
\end{align*}
From here one can see that each $|j \rangle$ after QFT becomes a superposition state of all basis states with equal probabilities (in this case the probability is equal to $\frac{1}{4}$). And because QFT is a unitary operator, if $\langle j | j'\rangle= 0$ (when $j \ne j'$), then $\langle j |QFT^{\dagger} QFT | j'\rangle= 0$, so the states generated by $QFT | j\rangle$ are different superposition states with equal probabilities that are orthogonal to each other.
Now three qubit case. I will write down only for three cases:
\begin{align*}
QFT &|000\rangle = QFT |0\rangle = \frac{1}{2^{\frac{3}{2}}} \sum_{k=0}^{7} e^{2 \pi i \frac{0 \cdot k}{2^n}} |k\rangle =
\\
&=\frac{1}{2^{\frac{3}{2}}}\big( |0\rangle + |1\rangle + |2\rangle + |3\rangle + |4\rangle + |5\rangle + |6\rangle + |7\rangle\big)
\\
QFT &|001\rangle = QFT |1\rangle = \frac{1}{2^{\frac{3}{2}}} \sum_{k=0}^{7} e^{2 \pi i \frac{1 \cdot k}{8}} |k\rangle =
\\
&=\frac{1}{2^{\frac{3}{2}}}\big( |0\rangle + e^{i \frac{\pi}{4}}|1\rangle + e^{i \frac{\pi}{2}}|2\rangle +e^{i \frac{3 \pi}{4}} |3\rangle + e^{i \pi}|4\rangle +e^{i \frac{5\pi}{4}} |5\rangle + e^{i \frac{3\pi}{2}}|6\rangle + e^{i \frac{7 \pi}{4}}|7\rangle\big)
\\
QFT &|111\rangle = QFT |7\rangle = \frac{1}{2^{\frac{3}{2}}} \sum_{k=0}^{7} e^{2 \pi i \frac{7 \cdot k}{8}} |k\rangle =
\\
&=\frac{1}{2^{\frac{3}{2}}}\big( |0\rangle + e^{i \frac{7 \pi}{4}}|1\rangle + e^{i \frac{3\pi}{2}}|2\rangle +e^{i \frac{5 \pi}{4}} |3\rangle + e^{i \pi}|4\rangle +e^{i \frac{3\pi}{4}} |5\rangle + e^{i \frac{\pi}{2}}|6\rangle + e^{i \frac{ \pi}{4}}|7\rangle\big)
\end{align*}
This time also $QFT |j\rangle$ generates superposition states with equal probabilities (note that $| \frac{e^{i\varphi}}{2^{\frac{3}{2}}}|^2 = \frac{1}{8}$ for any given $\varphi$ ) that are orthogonal to each other. The same logic works for arbitrary number of qubits $n$. $H$ can be regarded as one qubit QFT and note that $H |j \rangle$ ($j = 0,1$), in the same manner, also produces superposition states with equal probabilities that are orthogonal to each other.
If instead of computational basis $|j \rangle$ we apply QFT on an arbitrary superposition state $\sum_{j = 0}^{2^n -1} a_j |j\rangle$ things get slightly complicated:
$$QFT \sum_j a_j |j\rangle = \frac{1}{2^{\frac{n}{2}}} \sum_{l,k=0}^{2^n -1} e^{2 \pi i \frac{lk}{2^n}} | k \rangle \langle l | \sum_{j = 0}^{2^n -1} a_j |j\rangle = \frac{1}{2^{\frac{n}{2}}} \sum_{j,k=0}^{2^n -1} a_j e^{2 \pi i \frac{jk}{2^n}} | k \rangle $$
And the probability of measuring $|k \rangle$ is equal to:
$$p_k = \frac{1}{2^n} \left|\sum_{j = 0}^{2^n - 1} a_j e^{2 \pi i \frac{jk}{2^n}} \right|^2$$
As an example let's apply QFT on this Bell state $| \Phi^+ \rangle = \frac{1}{\sqrt{2}} \big(|00\rangle + |11\rangle \big) = \frac{1}{\sqrt{2}} \big(|0\rangle + |3\rangle \big)$:
$$QFT \frac{1}{\sqrt{2}} \big(|0\rangle + |3\rangle \big) = \frac{1}{2 \sqrt{2}} \big(2|0\rangle + (1 - i)|1\rangle + (1 + i)|3\rangle \big)$$
The probability of measuring $|0\rangle$ state is equal to $\frac{1}{2}$, but the probability of measuring $|1\rangle$ or $|3\rangle$ states are equal $\frac{1}{4}$. Also, note that the probability of measuring $|2\rangle$ state is zero in this case.