Answer: you can ignore the coating (assuming monotonic index of refraction); light does not exit if incident ray is beyond critical angle
Reasoning: Snell's Law states
$$n_1 \sin \theta_1 = n_2 \sin \theta_2,$$
where $n_1$ and $n_2$ are indexes of reflection within media $M_1$ and $M_2$, and
$\theta_1$ and $\theta_2$ are the angle between the normal and the light ray within the respective media.
I presume by "prism" you mean an interface $M_1$ to $M_2$, with $n_1 > n_2$.
Total internal reflections (TIR) occur when $\sin \theta_2 = \frac{n_1}{n_2} \sin\theta_1 > 1$, i.e., where transmission would be absurd since there is no real solution for $\theta_2$.
Now consider the case with $3$ media $M_1$, $M_2$, $M_3$ with parallel interfaces. Snell's law becomes
$$n_1 \sin \theta_1 = n_2 \sin \theta_2 = n_3 \sin \theta_3 = \text{constant}.$$
So it doesn't mater what $n_2$ is. As long as $\sin\theta_3 = \frac{\text{constant}}{n_3} > 1$, then $\theta_3$ has no real solution, and light will not enter $M_3$ (in fact, TIR may occur from the $M_1$-$M_2$ interface and never reach $M_3$ in the first place).
We can generalize the above to multiple layers. In the limit we would have a (assumed monotonic) gradient of index of fractions from $n_1$ to $n_2$, for each plane parallel to a common plane (from which the normal is defined).
Now consider a light ray coming from $M_1$ that would normally undergo TIR against $M_2$. For the gradient case, the light ray would bend and become increasing parallel to the plane, until it encounters a layer $M_i$ with $\frac{n_1}{n_i} \sin \theta_1 = 1$. Thereafter the ray bends back toward $M_1$, effectively undergoing TIR.
(Edit: updated "Answer" section part to match what the question is asking)
So at the boundary, what's really happening is: the frequency of light must remain the same (wave crests/troughs cannot exit any faster than they enter) and therefore the different speed of light changes the wavelength of the light.
Of a certain intensity of light, a proportion $T$ transmits into the new medium and a proportion $R$ exits. The barrier does not absorb energy, so $T + R = 1.$ It can only absorb momentum perpendicular to its surface, so there must be conservation of momentum in the other direction. Now for a light wave, the momentum is proportional to the inverse of the wavelength $\lambda$, so if $\theta$ is measured from the line perpendicular to the surface, momentum balance in this region means:$$\frac{1}{\lambda_1} ~ \sin \theta_1 = \frac{T}{\lambda_1}~\sin\theta_1 + \frac{R}{\lambda_2}~\sin\theta_2$$Conservation of momentum and energy therefore together give us Snell's law, since you can work out that if the frequency stays the same then $\lambda_1 / \lambda_2 = n_2 / n_1.$
You need another equation to successfully pin down the exact value of $R$ and see how it varies with respect to $\theta_1.$ This other equation is harder for me to explain, and it comes from the continuity of fields, which makes it polarization-dependent.
Typically light is a wave made of two fields, an electric and magnetic field, that oscillate perpendicular to each other, and perpendicular to the direction that the wave goes, too. So if light is going "up" then possibly the electric field is oscillating east-west while the magnetic field is oscillating north-south. The energy is contained in the square of the amplitude of these oscillations, and, if the wave is moving at a speed $v$, the magnetic field's amplitude is usually related to the electric field's amplitude by $E / v$. A full writeup of how exactly all this works is available here.
Here's one example. If the electric field points alongside the surface between the two media, then electric field continuity means that $E_i + E_r = E_t$. Defining the transmission amplitude $\tau = E_t / E_i$ and the reflection amplitude $\rho = E_r / E_i$, this means that $1 + \rho = \tau$. In general because it's the same medium, $R = \rho^2$ but $T \ne \tau^2$, so we need another equation. That's given by the magnetic field, $$\frac{1}{v_1} ~ (1 - \rho) ~ \cos\theta_1 = \frac 1 {v_2} ~\tau~\cos\theta_2, $$therefore
$$ (1 - \rho) ~ n_1 ~ \cos\theta_1 = (1 + \rho) ~ n_2 ~ \cos\theta_2$$or once you work it all out, $$R = \left({n_1 \cos\theta_1 - n_2\cos\theta_2 \over n_1 \cos\theta_1 + n_2\cos\theta_2 }\right)^2.$$Again, that's only true for this one polarization, but let's look at this formula.
First off, observe that this is actually really well-defined for $\theta_2 = \pi/2$, which is when we know $\theta_1$ is critical: it limits to 1. For all $\theta_1$ less than this, we get a reflection coefficient R which smoothly increases from 0 to 1 before abruptly having a "kink" at 1 (because Snell's law can no longer be satisfied beneath that point and so it stays $R = 1$). And this means that $T$ goes to 0.
So, as the transmitted ray gets closer and closer to being "alongside the surface", it also gets weaker and weaker in intensity. By the time the transmitted ray lies purely "alongside" the surface it has smoothly transitioned into being nonexistent.
Best Answer
The answer is YES. See diagram:
The key here is that you can write down the expressions for the angles $a_2, a_3, a_4$ in terms of the angles of the prism $\alpha, \beta, \gamma$ and the first angle $a_1$. Now $\alpha + \beta + \gamma = 180$ and you know you must have total internal reflection with the first three angles, but not with $a_4$.
$$a_2 = \gamma - a_1\\ a_3 = \beta - a_2\\ a_4 = \alpha - a_3$$
So expressing $a_4$ in terms of $a_1$ we get
$$\begin{align}\\ a_4 &= \alpha - (beta - (\gamma - a_1))\\ &= \alpha - \beta + \gamma - a_1\\ &= \alpha + \beta + \gamma - (a_1 + 2\beta)\\ &= 180 - (a_1 + 2\beta)\end{align}$$
Playing around a bit with these equations, I find that the optimal solution is obtained with
$$\alpha = 36\\ \beta = 72\\ \gamma = 72\\ a_1 = 36$$
In this case, you have $a_2 = 36, a_3 = 36, a_4 = 0$. In other words, as long as you can get total internal reflection with an angle of incidence of 36 degrees, you can do it - and as a bonus, you go "straight in" and come "straight out" - see below (approximately accurate):
This requires a refractive index just greater than 1.7 ($1/\sin(36)$ - definitely possible.
If you are willing to have the exit direction ($a_4$) be something other than zero, then you can improve on the above solution (make it possible with lower refractive index).