We know that giving a single qubit starting in the state $|0\rangle$, which is a state one can initialize very fast with high fidelity, then we can put it in the superposition state $|\psi \rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ by applying a Hadamard gate. That is,

$$H |0\rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$$

And we can do for $N$-qubit by applying each Hadmard gate to each qubit individually. That is,

\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}

You can do these $N$ operations of applying Hadamard gate to each qubit in parallel as applying a Hadamard to qubit 1 does not effect the state of qubit 2 and etc. You can see this parallelization in term of quantum circuit as well:

All these Hadamard gates can be and will be execute at the same time on the quantum processor.

Furthermore, each Hadamard gate can be executed pretty quick. In fact, if you execute a circuit with a Hadamard and do a measurement, the measurement process takes much much longer than the execution of Hadamard gate. See the pulse schedule for this particular circuit below: (The pink rectangular boxes indicate the measurement process)