Eviews gives the same p-value.
I think there is no "problem" here as the asymptotic distribution is
highly asymmetric around zero including a band of positive ADF values
but with very small probability area (and it is an one-sided test).
Actually you need a test statistic of 2.8 ( urcpval(2.8, 0, 1, 3) ) to
reach an asymptotic p-value of exactly 1 according to the code (and the
response surface estimated by MacKinnon (1996)).
P-value here (in the ADF test) is interpreted as the integral of the
distribution to the left of the ADF statistic. If you get an estimate of
ADF=0 the probability to the left is 99.62%, that is under the null
hypothesis you will observe an ADF statistic of 0 or smaller (<= than 0)
with 99.62 percent probability. Why should this be 1 or why should be 1
if the ADF statistic is positive?
The estimated parameter could be higher than one - although "slightly"
higher in relevant economic applications. For example you estimated
AR=1.00390523 so the asymptotic p-value is 0.99749368 or 99.74% (is
higher than 99.62 etc). Under the null that AR=1, the probability that
you will estimate (with constant trend etc that is with all
specification "correct") AR<=1.00390523 is 99.74%.
Now in the left tail assume that you get ADF=-3.5 so that eval
urcpval(-3.5, 0, 1, 3) gives 0.039304454 or under the null the
probability to "draw" an ADF less than or equal to -3.5 is "only" 3.93%.
At 5% (singificance) you reject the null etc.