I'll offer some hints as I think this is an exercise that you should get some practice with to really understand some of the tools being used.
As has already been mentioned a few times, it is probably easier to think of this space $X=\mathbb{R}^3\setminus T$ as the homotopy equivalent space $\mathbb{R}^3\setminus S^1$ where $S^1$ is an unknotted circle embedded in $\mathbb{R}^3$ (It's interesting to note that the qualifier 'unknotted' here is very important. If $S^1$ is knotted, then the fundamental group of its complement in $\mathbb{R}^3$ is known as the knot group of the knot and can be very different from the knot group of the unknot).
There are a few ways to tackle the problem now. Probably the easiest computationally (although perhaps hardest to visualise) is to note that $\mathbb{R}^3\setminus S^1$ is homotopy equivalent to the wedge sum of a $2$-sphere and a circle. That is $\mathbb{R}^3\setminus S^1\simeq S^2\vee S^1$. To see this, deformation retract $\mathbb{R}^3\setminus S^1$ on to the subspace $B\setminus S^1$ where $B$ is a solid ball containing $S^1$. Now, 'push' the negative space of the removed $S^1$ to the boundary of the ball so that we have the space $S^2\cup I$ where $I$ is an interval connecting the north and south pole of the sphere. You can now drag one end of the interval around to the other end so that it becomes a circle, giving $S^2\vee S^1$. The rest is just some simple applications of Seifert-van Kampen and knowing the fundamental groups of usual spaces.
Another approach would be to 'slice' $3$-space along a plane which also cuts the circle, leaving you with two half spaces, each with a removed interval whose ends are on the boundary of the half-spaces. If you 'thicken' these spaces up so that their intersection is open in $\mathbb{R}^3\setminus S^1$, then you can apply Seifert-van Kampen directly to this union. The only tricky part here is working out which elements in the fundamental group of the intersection correspond to the normal subgroup you will factor out of the free product.
One other approach might be to use the space that another user constructed in his answer.
EDIT that answer referenced was deleted so I will just say that it is the product of the space $S^1$ with $H\setminus \{p\}$ where $H$ is an open half plane, $p\in H$ and we view this space as a 'surface of revolution' in $\mathbb{R}^3$, which is homotopy equivalent to $(\mathbb{R}^3\setminus S^1)\setminus l$ where $l$ is an infinite line which passes through $S^1$.
If you then consider that the union of this space with an open tubular neighbourhood of $l$ is exactly the space you are considering, then another application of Seifert-van Kampen should give you the fundamental group of your space readily (I haven't written this down so I'm not sure if there will be any difficulties in this decomposition, but I can't see why there would be).
I call $A$ the space that you are talking about. $A$ is of the form $A=\Lambda-T$ where $\Lambda$ is the original donut, and $T$ is the "tunnel" (it is a specific subspace of $\Lambda$ homeomorphic to a torus). I claim that $A$ deforms retract onto a subset $X$ such that
$$X\simeq \Bbb T^2 \bigsqcup \Bbb M^2/\sim$$
where $\Bbb M^2$ is the mobius band and $\sim$ is the relation attaching the boundary of $\Bbb M^2$ along one "canonical generator" of the fundamental group of the torus $\Bbb T^2$. The space $X$ is just "a mobius band which has his usual boundary replaced by a torus". To see that, I did the following drawing:
In my drawing, the big torus $\Lambda$ is in green, the tunnel $T$ created by the worm is in red and $X$ is the union of $T$ and the hatched grey mobius band. The mobius band is created by drawing the lines as in the picture.
In order to understand that $A$ deforms retract onto $X$, think of it the way around. If you take a neighborhood of $X$ in $A$, it looks just like $A$.
The only thing left to do is to compute the fundamental group of $X$. The space $X$ can be seen as follows, with the identifications by colors:
I don't know what is the best way to compute $\pi_1(X)$. Here is two ways that I can think of:
You can use the "classic trick" that consists in doing a hole in the space and then using Van Kampen to get the fundamental group of the original space. To be more precise, take a disk $D^2\subset X$ and $p\in D^2$. Then if $\gamma$ is a generator of $D^2-\{p\}$ and if $i:D^2-\{p\}\to X-\{p\}$ is the inclusion, the theorem of Van-Kampen implies
$$\pi_1(X)\simeq\pi_1(X-\{p\})/_{\langle i_*(\gamma)\rangle}$$
More interestingly, in the picture we see $X$ as the quotient space
$$X= Y\times [0,1]/_{Y\times 0 \stackrel{\varphi}{\sim} Y\times 1}$$ where $Y$ is two circles joined by a straight line and $\varphi$ is a homeomorphism that "turns $Y$ upside down". This is perfect to use the quotient version of Van Kampen. This version is not very famous, and the only reference I have for it is in French (but it's actually a great reference if you can understand French, there is a very comprehensive video going with it!).
I'm sure there is some alternative ways to compute $\pi_1(X)$. Anyway using the first method I found
$$\pi_1(X)\simeq \langle a,b~\vert~ bab^{-2}a^{-1}b=1\rangle.$$
In my opinion, this was not obvious at all, I can add details/drawings if needed. But I feel that you should give it a try using Van Kampen now that the situation is more clear!
I hope this helps!
Edit: I added some detail changed the names of the spaces (the previous names were implicitly saying that $\Lambda$ and $T$ had dimension 2 instead of 3)
Edit 2: As smartly suggested by Kyle Miller in the comments, there is an easy way to compute $\pi_1(X)$. If $a$ and $b$ are two generators of the torus such that $X$ is obtained by identifying the boundary of the mobius band with $a$, and if $c$ is a (well chosen) generator of the fundamental group of the mobius band, the theorem of Van Kampen gives
$$\pi_1(X)\simeq \langle a,b,c~\vert~ ba=ab,~c^2=a\rangle,$$
which can be rewritten
$$\pi_1(X)\simeq \langle b,c~\vert~ bc^2=c^2b \rangle,$$
which is the same as in the other presentation.
Best Answer
This answer will be a bit incomplete, but I think you can work out the full details from it (feedback is welcome).
Consider you have the solid torus with the torus shaped hole winding ally around it twice, and call this space $X$, then you want to calculate $\pi_1(Z)$, where $\frac{X}{Y_1\sim_f Y_2}$, $Y_1,Y_2$ are the connected components of $\partial X$ and $f:Y_1\to Y_2$ is a homeomorphism. Now, with the notation out of the way, cover $Z$ with two open sets (Van Kampen in the making) $U,V$, where $U = \mathrm{Int} X$, and $V$ is the quotient (by the relation $\sim_f$) of an open neighborhood of $\partial X$; further, since $U\cap V$ is not connected, we are gonna have to use the groupoid version of the Van Kampen theorem.
So choose two points in $U \cap V$, which I will call $0$ and $1$ because I don't want to use more letters, and we calculate the fundamental groupoids of $U,V$ and $U\cap V$, with respect to the set of basepoints $\{0,1\}$. Each of these groupoids has two elements, and all we need to determine is the automorphism group of these two elements (corresponding to $0$ and $1$) and the morphisms between them (if they exist).
For $\pi_1(U\cap V,\{0,1\})$, each element only has the fundamental group of the torus (\mathbb Z \oplus \mathbb Z) as it's automorphism group, and since $0$ and $1$ lie in different components of $U\cap V$ there are no morphisms (which correspond to paths) between them.
For $\pi_1(V,\{0,1\})$, you can see that since $V$ deformation-retracts to the boundary of $X$ under the identification $\sim_f$. Further, you get that, once again, under this identification the boundary of $X$ is homeomorphic to the torus, and so $\hom(0,0) = \hom(1,1) = \pi_1(V) = \mathbb Z \times \mathbb Z$. Also, by choosing the base-points $0$ and $1$ such that $1 \in r^{-1}(f\circ r (0))$, one can see that all elements of $\hom(0,1)$ are of the form $\circ \gamma_{0,1} \circ k_0$, where $\gamma_{0,1}$ is a path from $0$ to $1$ and $k_0$ is an element of $\hom(0,0)$ (note that $\gamma_{0,1}\circ k_0 = h(k_0)\circ\gamma_{0,1}$ where $h$ is the "identity" $\hom(0,0) = \hom(1,1)$.
Finally (we still have to see what are the maps, but, we're almost there) for $\pi_1(U,\{0,1\})$, we can see that $U$ deformation-retracts to a torus with a Möbius band inside it, and so has its fundamental group given by $\pi_1(X) = \frac{\mathbb Za + \mathbb Zb + \mathbb Zc}{2a + b \sim 2c}$ (there is a mistake in my comment before, sorry about that). And once again, $\hom(0,0) = \hom(1,1) = \pi_1(X)$ and $\hom(0,1)$ has elements of the form $\delta_{0,1}\circ k_0$.
Now, all that is left is to assemble the pieces by choosing a point, (either $0$ or $1$, I will choose $0$).
The map $\pi_1(U\cap V,0)\to \pi_1(V,0)$ is the identitiy
The map $\pi_1(U\cap V,0)\to \pi_1(U,0)$ takes one generator to $a$ and the other generator to $b$, as per the representation given above (note that I am assuming $0$ is the point "close to the outer boundary of $X$")
Finally, we need to consider the group elements that arise from composing elemnts of $\hom(0,1)$ with elements of $\hom(1,0)$, these will have the form $\delta^{-1}\circ\gamma$ which is not homotopy equivalent to any of the other generators.
So, we get $\pi_1(Z) = \frac{\mathbb Za + \mathbb Zb + \mathbb Zc + \mathbb Z(\delta^{-1}\gamma)}{2a + b \sim 2c}$