To perform such a calculation, we will use the flat Minkowski space-time of special relativity. Assuming that the traveller does not come sufficiently close to any massive body during the trip.
Now, in order to perform this calculation relativistically (assuming you want to include the effects of a changing Lorentz factor and associated acceleration) we must first obtain/derive an expression for how the Lorentz factor of a moving object transforms. This will allow us to derive the required relativistic expression for the objects proper-acceleration.
So, let us consider two inertial frames S (the 'stay-at-home observer) and S' (the traveller) in 'standard configuration' (that is, assuming that S' is moving in the positive x-direction with speed $v$). Let $\mathbf{u} = (u_{1}, u_{2}, u_{3})^{\mathsf{T}}$ be the instantaneous vector of velocity in S of the traveller. We now wish to find the velocity and $\mathbf{u}' = (u'_{1}, u'_{2}, u'_{3})^{\mathsf{T}}$ of the traveler in frame S'. We can define
$$\mathbf{u} = (\mathrm{d}x/\mathrm{d}t, \mathrm{d}y/\mathrm{d}t, \mathrm{d}z/\mathrm{d}t))^{\mathsf{T}},$$
$$\mathbf{u}' = (\mathrm{d}x'/\mathrm{d}t', \mathrm{d}y'/\mathrm{d}t', \mathrm{d}z'/\mathrm{d}t'))^{\mathsf{T}}.$$
From this definition and the fact that the two frames are in the 'standard configuration', we can immediately write the well know velocity transformation formulae (without derivation):
$$u'_{1} = \frac{u_{1} - v}{1 - u_{1}v/c^{2}}, \; u'_{2} = \frac{u_{2}}{\gamma(1 - u_{1}v/c^{2})}, \; u'_{3} = \frac{u_{3}}{\gamma(1 - u_{1}v/c^{2})}. $$
No assumptions about uniformity were made here and these formulae apply equally to the instantaneous velocity in a non-uniform motion.
Let us now write $u = (u_{1}^{2} + u_{2}^{2} + u_{3}^{2})^{\frac{1}{2}}$ and $u' = ({u'}_{1}^{2} + {u'}_{2}^{2} + {u'}_{3}^{2})^{\frac{1}{2}}$ for the magnitudes of the corresponding velocities in S and S'. Now let us choose the signature of our metric tensor $g_{\mu \nu}$ of our Minkowski space-time so that for our two inertial frames we can write
$$c^{2}\mathrm{d}t'^{2} - \mathrm{d}x'^{2} - \mathrm{d}y'^{2} - \mathrm{d}z'^{2} = c^{2}\mathrm{d}t^{2} - \mathrm{d}x^{2} - \mathrm{d}y^{2} - \mathrm{d}z^{2}.\;(\mathrm{A})$$
In our 'standard configuration' the Lorentz transformations for our coordinates are given by
$$\mathrm{d}x' = \gamma(\mathrm{d}x - v\mathrm{d}t), \; \mathrm{d}y' = \mathrm{d}y, \; \mathrm{d}z' = \mathrm{d}z, \; \mathrm{d}t' = \gamma(\mathrm{d}t - v\mathrm{d}x/c^{2}).\;(\mathrm{B})$$
Now, factoring out $\mathrm{d}t'^{2}$ and $\mathrm{d}t^{2}$ form the LHS and RHS of (A), respectivley and using (B) we can write
$$\mathrm{d}t^{2} (c^{2} - u^{2}) = \mathrm{d}t'^{2}(c^{2} - u'^{2}) = \mathrm{d}t^{2}\gamma^{2}(v)(1 - u_{1}v/c^{2})^{2}(c^{2} - u'^{2}).\;(\mathrm{C})$$
Now, cancelling $\mathrm{d}t^{2}$ from the above we can now obtain the following transformation for $u^{2}$, the squared magnitude of our traveller's velocity:
$$c^{2} - u'^{2} = \frac{c^{2}(c^{2} - u^{2})(c^{2} - v^{2})}{(c^{2} - u_{1}v)^{2}}.$$
Note here $u_{1}v = \mathbf{u}.\mathbf{v}$ so that the RHS is actually symmetric in $\mathbf{u}$ and $\mathbf{v}$ - meaning that this holds for any two subliminal 3-velocities. Now, rewriting the above interms of $\gamma(u)$ and $\gamma(u')$, with some work we get the following useful relations
$$\frac{\gamma(u')}{\gamma(u)} = \gamma(v)(1 - \frac{u_{1}v}{c^{2}})$$
This expression shows how the Lorentz factor of a moving object transform (for +ive $v$). We can now use this to get an expression for our proper-acceleration.
Now using the rapidity function we can simplify the following derivation (in a big way!). The rapidity function $\phi(u)$ can be written as
$$\phi(u) = \tanh^{-1} (\frac{u}{c}), $$
which allows the velocity addition formula to be rewritten in the remarkably simple form
$$\phi(u) = \phi(v) + \phi(u'),$$
now differentiating with respect to (WRT) $t$ yields
$$\frac{\mathrm{d}}{\mathrm{d}t}\phi(u) = \frac{\mathrm{d}}{\mathrm{d}t'}\phi(u')\frac{\mathrm{d}t'}{\mathrm{d}t}.\;(\mathrm{D})$$
Which can be written as
$$\frac{\mathrm{d}}{\mathrm{d}t}\phi(u) = \frac{1}{c}\gamma^{2}(u)\frac{\mathrm{d}u}{\mathrm{d}t},$$
and from (C) above we can write $\mathrm{d}t'/\mathrm{d}t = \gamma(u')/\gamma(u)$. Substituting this and the equation above (and its primed version) into the above expression for $\phi(u)$ we can write the desired acceleration transformation formula (hoping I have made no mistakes! :])
$$\gamma^{3}(u')\frac{\mathrm{d}u'}{\mathrm{d}t'} = \gamma^{3}(u)\frac{\mathrm{d}u}{\mathrm{d}t}.$$
Now if we define the proper acceleration $\alpha$ (say), as that which is measured in our travellers rest-frame S', we find on setting $u' = 0$ and $\mathrm{d}u'/\mathrm{d}t' = \alpha$, using our acceleration transformation equation we get
$$\alpha = \gamma^{3}(u)\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}[\gamma(u)u].$$
This proper acceleration $\alpha$ is exactly the push we feel in an accelerating rocket. Now finally, in our case of interest, that of rectilinear motion with constant proper acceleration $\alpha$. We can integrate the above equation once, choosing $t = 0$ when $u = 0$
$$\alpha t = \gamma(u) u. \; (\mathrm{E})$$
Square this, solve for $u$ and integrate again with the same initial conditions give us the following equation of motion
$$x^{2} - c^{2}t^{2} = c^{4}/\alpha^{2}.$$
hence why motion with constant proper accelleration is called hyperbolic!
We can now solve your question. It is likely that at 20g for half the distance we will be well beyond the speed of light. Let us take the version of the derived relativistic equation of motion above for the frame S. Now, setting the distance as 2.125 light years with an acceleration of 20g, we can work out the time taken to reach the halfway point (using the relativistic equation of motion above), from the home observers reference frame which turns out to be 1531 days (or 4.19 years). This motion will be symmetric so the time taken for the entire trip in frame S (taking into account full relativistic motion!) will be 3062 days (or 8.39 years).
Now for the time taken as measured in frame S'... I will let you work that out! It is not as simple as using a Lorentz transform on the total time taken in this case; as we have seen that the Lorentz factor will change for and accelerating body.
As for the maximum speed I will also leave this as an exercise - I have purposely missed out the step where we derive equation for $u$. You can get $u$ from (E), and work out the max speed accordingly.
You will also notice that in the Newtonian calculation, the time taken to get to the half-way point is 166 days. This is because the speed of light is reached in 17.69 days at a distance of 212 light minutes; giving a speed at the half-way mark of a whopping 9.38c! The relativist calculation reflects the limit of c in the calculation.
I hope you enjoy reading this as much as I did going through it. You can tell I am off work!
All the best.
I'm going to first address some misconceptions you seem to have and then I will get to answering your question. Now, as stated in the comments, @Pulsar did a very thorough job of answering this question in another post. But I read through that answer and it's a bit technical. I already knew the stuff, so it made sense to me, but I can see how someone unfamiliar with the material might find that to be more like science gobbledegook than a clear and helpful explanation.
So first things first, we think the universe is about 13.8 billion years old. Second, the universe is expanding, which means the size of the observable universe has expanded to larger than the distance light travels in that time. The diameter of the observable universe is thought to be about 93 billion light years and no light from us will ever reach this edge again (if we're right in our model of the universe).
Next, I'll get slightly more technical. A light year can not be shorter, longer, or the same length as an Earth year. One is a unit of distance, the other is a unit of time. That is like saying a metre is shorter than an hour. A light year has a rigidly defined distance length. Furthermore, a photon has no reference frame in which to define a length or a time. There is no light year that pertains to light itself.
Okay now, let's get even more technical. The universe is expanding. We know this. Because of the expansion and because of this wonderful and not at all magical thing called the Cosmic Microwave Background (CMB), we can identify a frame of reference of the universe that is at rest with respect to what we assume is the stationary background of spacetime (once you factor out the expansion). We call this the comoving frame. Anything at rest in the comoving frame sees the CMB as the same from all directions and the only reason things in the comoving frame move away from each other is because of expansion. The reason we like this frame is because it is a universally discoverable frame. Aliens from a distant galaxy would be able to find this frame and agree on what observers in it would see. Having said that, the age of the universe is calculated in this frame. The detailed calculations are all in @Pulsar's post, but when we say the universe is 13.8 billion years old, we have figured out the number of Earth years in the comoving frame since the Big Bang era began. That means that no matter who is trying to find out the age of the universe, they'll be able to find the comoving frame and figure out the age in that frame and get the same value.
Best Answer
Your question can be translated into "if right now we would send a powerful omnidirectional light pulse from earth into space, would there be galaxies that never see this light pulse?"
The answer is "yes". Due to the accelerated expansion of the universe, as described by the lambda-CDM model, only galaxies currently less than about 16 billion light years (the difference between the cosmological event horizon and the current distance to the particle horizon) away from us will at some time observe the light pulse.
A nice visual representation of this can be found in figure 1 of this publication.