As Karsus points out, you only ever measure energy differences during transitions between different levels, and it is hard for these to put too much energy into the translational motion. To get a quantitative answer, consider that a photon can only give about $\hbar k$ of momentum to the atom, and this will be weighted down by the total mass ($\approx$ the proton mass), which is far bigger than the reduced mass ($\approx$ the electron mass) which governs the spectrum. The total kinetic energy difference will therefore be of the order of $$\frac{(\hbar k)^2}{2m_p}=\frac{\hbar^2\omega^2}{2m_pc^2}=\frac{(\Delta E)^2}{2m_pc^2}$$ and is therefore smaller than $\Delta E$ by a factor of $\Delta E/2m_pc^2\lesssim7\times10^{-9}$. This effect will therefore cause a broadening of the transition lines by that amount and it is called the Doppler limit on the precision of the transition frequency. This is of course independent of how many atoms you're addressing.

However, the translational motion does affect the energies you observe by Doppler shifting them, so that if you're observing a single atom the line will move to higher or lower frequency by a factor of $\sim v/c$. If you have many atoms moving in many different directions at different velocities, then each will radiate at its own natural frequency but you will observe a bunch of different Doppler shifts, and therefore the line will be broadened by a factor of $\sim v_\textrm{th}/c$ where $v_\textrm{th}$ is the thermal velocity given by $m v_\textrm{th}^2\approx k_B T$. This is Dopppler broadening and it is dominant if you do nothing to bring it down. It will certainly be dominant in a high-temperature environment like an arc lamp.

As Karsus points out, you only ever measure energy differences during transitions between different levels, and it is hard for these to put too much energy into the translational motion. To get a quantitative answer, consider that a photon can only give about $\hbar k$ of momentum to the atom, and this will be weighted down by the total mass ($\approx$ the proton mass), which is far bigger than the reduced mass ($\approx$ the electron mass) which governs the spectrum. The total kinetic energy difference will therefore be of the order of $$\frac{(\hbar k)^2}{2m_p}=\frac{\hbar^2\omega^2}{2m_pc^2}=\frac{(\Delta E)^2}{2m_pc^2}$$ and is therefore smaller than $\Delta E$ by a factor of $\Delta E/2m_pc^2\lesssim7\times10^{-9}$. This effect will therefore cause a broadening of the transition lines by that amount. This is of course independent of how many atoms you're addressing.

However, the translational motion does affect the energies you observe by Doppler shifting them, so that if you're observing a single atom the line will move to higher or lower frequency by a factor of $\sim v/c$. If you have many atoms moving in many different directions at different velocities, then each will radiate at its own natural frequency but you will observe a bunch of different Doppler shifts, and therefore the line will be broadened by a factor of $\sim v_\textrm{th}/c$ where $v_\textrm{th}$ is the thermal velocity given by $m v_\textrm{th}^2\approx k_B T$. This is Dopppler broadening and it is dominant if you do nothing to bring it down. It will certainly be dominant in a high-temperature environment like an arc lamp.