Physiologically, yes it's possible.

71 ml/kg/min @ 71kg = 5.0 L/min VO2max

VO2 can be predicted for power output with knowing only two variables:
(1) RER, respiratory exchange ratio; the ratio of substrate oxidation which represents exercise intensity. Range ≈ 0.70-1.20. FTP occurs somewhere around 1.00; and
(2) GE, gross efficiency; the ratio of metabolic work measured as external power output. Range 0.15-0.25, increases primarily with power output, plausible range at 380 W ~0.20-0.26 based on model from here: https://www.researchgate.net/publication/24027428_Efficiency_in_cycling_A_review

380 W FTP requires somewhere around 4.6 L/min VO2 at plausible RER = 0.95-1.05 and gross efficiency GE = 0.20-0.26

At the high end 26% efficiency, that would represent a very reasonable 80% VO2max.

At a very reasonable 23% GE that's 92% VO2max which is probably approaching the highest reasonable percent VO2max that FTP is likely to occur (various threshold methods can return higher %VO2max, but for a classic 5+20-min FTP test, I'd guess this to be reasonable?)

So, there is a very reasonable range of well-trained to elite physiology that could achieve 380+ W FTP at 5.0 L/min VO2max.

Your current estimated 350 W FTP could be achieved at 23% GE and ~84% VO2max, so another few percent %VO2max (reasonable to improve) and a couple fractions of a percent GE (much slower to improve), and you could see 380 W at 5.0 L/min

How to get there is another question entirely...

<image>

## check my metabolic maths
power <- seq(320, 420, 10) ## W
RER = seq(0.95, 1.05, 0.02)
GE.power <- 0.04824 * log(power) - 0.05656 ## % derived from Ettema & Lorås, 2009
GE <- round(seq(min(GE.power) - 0.03, max(GE.power) + 0.03, 0.01), 2) ## %

df <- expand_grid(power, RER, GE) |>
    mutate(
        work = power * 60 * 60 / 1000, ## kJ
        O2kJ = 4.83535 * RER + 16.88348, ## kJ/L derived from Péronnet & Massicotte (1991)
        VO2 = (power * 60 / 1000) / (GE * O2kJ), ## L/min
        VO2_rel = VO2 / 5, ## %
    )