Assume an electron which is moving very slowly and we observe it with a distance uncertainty of say Δx=1×10−13 m

In QM, particles don't have velocities in the normal sense of the word. Velocity is an observable, and thus is represented by an operator applied to a quantum state. Talking of a particle's "velocity" implies that the particle has a definite velocity (i.e. is in an eigenstate of the velocity operator) or, at the very least, its state has a small spread in velocity space. As you calculated, an electron with such a small $\Delta x$ would have such a massive $\Delta p$ that it cannot be said to have anything close to a well-defined velocity.

If an electron is moving close to $c$, then it will traverse $10^{-13}m$ in ~$3*10^{-22}$ seconds. According to a cursory web search that I performed, the highest time precision ever recorded is $10^{-21}s$. https://www.smithsonianmag.com/smart-news/physicists-record-smallest-slice-time-yet-180961085/ So it's not possible to measure an electron over short enough time period for it to be confined within a region of $10^{-13}m$.

That is not to say that it is not legitimate to ask about a purely hypothetical, completely unmeasurable scenario in which over a period of less than a zeptosecond, an electron has $\Delta x = 10^{-13}m$. I just thought it should be pointed out that this is a physically unrealistic situation.

As for this apparently resutling in $\Delta v > c$, as Valter Moretti says, your calculation is based on $p = mv$, and if $m$ is taken to be the rest mass $m_0$, then this is valid only for small $v$ (relative to $c$). However, I don't think Valter Moretti's further calculations are valid. The $\Delta p$ in the uncertainty is not the range of $p$, although this interpretation is a good enough approximation to be a good intuition when the priciple is being introduced. Rather, $p$ the standard deviation of $p$: $\sqrt {<\phi^* |p \phi>^2-<\phi^* |p^2 \phi>}$. Since $p$ is a nonlinear function of $v$, we can't calculate an exact value of $\Delta v$ in terms of $\Delta p$ without knowing the exact $\phi$.

Assume an electron which is moving very slowly and we observe it with a distance uncertainty of say Δx=1×10−13 m

In QM, particles don't have velocities in the normal sense of the word. Velocity is an observable, and thus is represented by an operator applied to a quantum state. Talking of a particle's "velocity" implies that the particle has a definite velocity (i.e. is in an eigenstate of the velocity operator) or, at the very least, its state has a small spread in velocity space. As you calculated, an electron with such a small $\Delta x$ would have such a massive $\Delta p$ that it cannot be said to have anything close to a well-defined velocity.

If an electron is moving close to $c$, then it will traverse $10^{-13}m$ in ~$3*10^{-22}$ seconds. According to a cursory web search that I performed, the highest time precision ever recorded is $10^{-21}s$. https://www.smithsonianmag.com/smart-news/physicists-record-smallest-slice-time-yet-180961085/ So it's not possible to measure an electron over short enough time period for it to be confined within a region of $10^{-13}m$.

That is not to say that it is not legitimate to ask about a purely hypothetical, completely unmeasurable scenario in which over a period of less than a zeptosecond, an electron has $\Delta x = 10^{-13}m$. I just thought it should be pointed out that this is a physically unrealistic situation.

As for this apparently resutling in $\Delta v > c$, as Valter Moretti says, your calculation is based on $p = mv$, and if $m$ is taken to be the rest mass $m_0$, then this is valid only for small $v$ (relative to $c$). However, I don't think Valter Moretti's further calculations are valid. The $\Delta p$ in the uncertainty is not the range of $p$, although this interpretation is a good enough approximation to be a good intuition when the priciple is being introduced. Rather, $\Delta p$ is the standard deviation of $p$: $\sqrt {<\phi^* |p \phi>^2-<\phi^* |p^2 \phi>}$. Since $p$ is a nonlinear function of $v$, we can't calculate an exact value of $\Delta v$ in terms of $\Delta p$ without knowing the exact $\phi$.