Noise PSD and sampling rate relation
Important Information : Sampling at $f_{s}$ will map $[-\frac {f_{s}}{2}, \frac {f_{s}}{2}]$ to digital frequency $\omega=[-\pi, \pi]$, and similarly sampling at $2f_{s}$ will map $[-f_{s}, f_{s}]$ to digital frequency $\omega=[-\pi, \pi]$. Also, we need to look only at digital frequency $\omega = [-\pi, \pi]$ as the digital spectrum is a $2\pi$-periodic function.

Now, lets look at discrete White Gaussian Noise Samples which you have generated.
$$n_{1} = N(0,\sigma^{2}), n_{2} = N(0, \sigma^{2})$$

This means auto-correlation function for both $n_1$ and $n_2$ will be:
$$r[k] = \sigma^{2}\delta[k]$$
and hence, the PSD of $n_1$ and $n_2$ in digital domain will be :
$$P(e^{j\omega}) = \sigma^2, \omega \in [-\pi, \pi]$$
and hence total Noise Power will be :
$$T_{NoisePower} = 2\pi.\sigma^2$$

Basically, once you have discrete Noise Samples for a White Gaussian Process, then autocorrelation function is fixed and hence the PSD in digital domain is fixed too. Sampling frequency does not matter for determining Noise Power.

Though, oversampling can help in increasing SNR of the received Signal. Think about it!

Hint: Oversampling followed by filtering!!