One-g is large enough to make this easier: Times are short.
Boosting straight out from Earth (taking $g = 10 m/s^2$ and ignoring the $0.006 m/s^2$ acceleration toward the Sun) takes
$d = (230 \times 10^6 \times 10^3 - 150 \times 10^6 \times 10^3)/2 = 1/2 g (t/2)^2$
$t = 2 \times \sqrt{ {2 \times 40 \times 10^ 6 \times 10^3} \over {g}}$
For a 1 g acceleration, that's $180 \times 10^3 s$ or about 50 hours.
Turning that acceleration to also change the speed along the orbit by $6 km/s$ only requires a $ 6,000 / 180,000 = 0.033 m/s^2$ acceleration along the orbit, or changing the angle of the thrust by just about 3 mrad. And that, in turn, only reduces the radial acceleration by $g (1 - cos(0.0033)) = .05 mm/s^2$, much less than the acceleration due to the Sun.
So I propose that the fastest way to get to Mars at 1g is to thrust straight toward the rendezvous point and orbital velocity you want, and turn over when half way there.
When you do that at 1 g, you build up outward speed so fast that the initial along-orbit speed is swamped. It takes less than an hour to build up an outward velocity larger than the initial orbital velocity! You really don't see the curves at the end of the sketch above:
(small custom python simulation; mars is red, earth is blue) When you draw the rocket path with one-hour dots, you can see it's really moving at turnover:
Finally, to answer the question about the fastest possible trip given g tolerance: That scales (in hours and g's) as $50/ \sqrt{g_{max}}$. If you can handle 4g for an extended time, you can be at Mars in about a day!
(Edit to include some comments)
Is this the fastest strategy?
In the simplified case (1) transfer between two straight-line moving objects (2) with no overall gravitational effects from the Sun and (3) no local gravitational effects from the starting and ending at planets, perhaps in orbit, then it's clear that a "go straight, turnover at the appropriate moment to make speed" is fastest. Not a single Gal of acceleration wasted!
How important are those assumptions?
For accelerations large enough that times are short (10 days?), curvature (1) and the Sun's 0.006g acceleration (2) don't change the strategy. You're still going as fast as you can to get to the place you need to be. For exact arrival, you've have to include it in calculations of direction and time, but this is a little bit like the difference between celestial navigation for an airplane and a sailboat: The boat moves slowly enough that navigation to a mile or two accuracy matters; much less accuracy is needed for the plane because it can get within 10's of miles and just home on the destination without major inconvenience.
Questions about departure and arrival details are more complicated; this Answer has ignored them. Leaving when your orbit is pointed at the destination is faster than leaving when it's pointed away. The initial and final orbits might be inclined, the planet orbits are askew, etc. But at large accelerations (i.e. anything near "survivability limits of humans") these are just computational inconveniences: A constant 1g flight starts off slower than many transfer orbit insertions (through you have to start in orbit; you can't leave earth's surface at 1g) but then you keep going all day, ending that day going 10X as fast as LEO in whatever direction you want.
