Do you mean mapping the state $|0\rangle^{\otimes n} \to \dfrac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1} |i\rangle $ ?

If that is the case then you can just apply $H^{\otimes n}$ to the state $|0\rangle^{\otimes n}$. That is, you apply a Hadamard gate to each of the qubit.

The reason for this is $H |0\rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ and so

\begin{align} \overbrace{H|0\rangle \otimes H|0\rangle \otimes \cdots \otimes H|0\rangle}^{n \ \textrm{times}} &= \overbrace{ \bigg( \dfrac{|0\rangle +|1\rangle}{\sqrt{2} } \bigg)\otimes \bigg( \dfrac{|0\rangle +|1\rangle}{\sqrt{2} } \bigg) \otimes \cdots \otimes \bigg( \dfrac{|0\rangle +|1\rangle}{\sqrt{2} } \bigg) }^{n \ \textrm{times}} \\
&= \dfrac{1}{\sqrt{2^{n}}}\big( \overbrace{ |00\cdots0\rangle + |00\cdots1\rangle + \cdots + |11\cdots 1\rangle }^{2^n \ \textrm{terms} } \big)\\
&= \dfrac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1} |i\rangle
\end{align}

5Initialize them all to $\vert 0\rangle$, and Hadamard them all individually. – Mark S – 2020-12-03T01:17:09.970

@Mark S Thanks!!! – Zhengrong – 2020-12-03T01:21:50.253