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).