Here is the plot of the speed of the Voyager (by its distance from the Sun):

Source: this question
We can assume an about $\rm{20 \frac{km}{s}}$ average speed, what is ${\rm \approx \frac{2}{30000} c}$.
For such low velocities, we can use the approximative time dilation formula: $\rm{\Delta t'=\Delta t \frac{v^2}{2c^2}}$. That gives $\approx 4.44 \cdot 10^{-8}$, meaning a special relativistic time dilation of $\approx$ 1.4 seconds in a year.
Voyager 1 was launched in 1977.09.05, about 45 years ago. Going back on the same path would make an about 90 years travel time, giving the result of about $\rm{\underline{\underline{2 min}}}$ .
We have also general relativistic effects which should be counted, although they do not affect the result too much. The general relativistic time dilation (time slowing due to gravity) caused by the Sun at Jupiter orbit is:
$$\Delta t'=\Delta t \sqrt{1-\frac{v_e^2}{c^2}}$$
$v_e$ is the escape velocity. The orbital speed of the Jupiter is $\rm{\approx 13.6 \frac{km}{s}}$, thus $\rm{v_e \approx 19.3 \frac{km}{s}}$. But the Voyager has spent most if its time far more away as the Jupiter. It might cause maybe some tens of seconds dilation in the result, but not more.
A lesser approximative calculation could be done based on more detailed public data, but it would be a big work and would not affect significantly the result.