Extending to $d$-dimensions
So I had enough time to procrastinate today and I extended this problem to $d$ dimensions. I would appreciate if someone could read through the answer and suggest any simplification to final answer if possible. Do the final numbers $u_n$ constitute a nice well known sequence? Further, what other interesting problems arise out this ant and the cube problem. I think another nice problem is the expected cover time. Thanks
Working out explicitly for $d=3$ gave some nice hints and a good picture of what is going on. We use these to extend to any d-dimension.
The first thing is to define a cube in $d$-dimensions. A $\mathbf{d}$-dimensional cube is a set of vertices of the form $(i_1,i_2,\ldots,i_d)$ where $i_k \in \{0,1\}$. There is an edge between vertex $(i_1,i_2,\ldots,i_d)$ and vertex $(j_1,j_2,\ldots,j_d)$ if $\exists l$ such that $\left | i_l - j_l \right| = 1$ and $i_k = j_k$, $\forall k \neq l$.
Two vertices are said to be adjacent if they share an edge. It is easy to note that every vertex shares an edge with $d$ vertices. (Consider a vertex $(i_1,i_2,\ldots,i_d)$. Choose any $i_k$ and replace $i_k$ by $1-i_k$. This gives an adjacent vertex and hence there are $d$ adjacent vertices as $k$ can be any number from $1$ to $d$). Hence, the probability that an ant at a vertex will choose an edge is $\frac1{d}$.
For every vertex, $(i_1,i_2,\ldots,i_d)$ let $s((i_1,i_2,\ldots,i_d)) = \displaystyle \sum_{k=1}^d i_k$. Note that $s$ can take values from $0$ to $d$. There are $\binom{d}{r}$ vertices such that $s((i_1,i_2,\ldots,i_d)) = r$.
Let $v_{(i_1,i_2,\ldots,i_d)}$ denote the expected number of steps taken from $(i_1,i_2,\ldots,i_d)$ to reach $(1,1,\ldots,1)$
Let $S_r = \{ (i_1,i_2,\ldots,i_d) \in \{0,1\}^d: s((i_1,i_2,\ldots,i_d)) = r\}$
Claim: Consider two vertices say $a,b \in S_r$. Then $v_a = v_b$. The argument follows easily from symmetry. It can also be seen from writing down the equations and noting that the equations for $a$ and $b$ are symmetrical.
Further note that if $a \in S_r$, with $0 < r < d$, then any adjacent vertex of $a$ must be in $S_{r-1}$ or $S_{r+1}$. Any adjacent vertex of $(0,0,\ldots,0)$ belongs to $S_1$ and any adjacent vertex of $(1,1,\ldots,1)$ belongs to $S_{d-1}$. In fact, for any $a \in S_r$, $r$ adjacent vertices $\in S_{r-1}$ and $d-r$ adjacent vertices $\in S_{r+1}$.
Let $u_r$ denote the expected number of steps from any vertex $\in S_r$ to reach $(1,1,\ldots,1)$. For $r \in \{1,2,\ldots,d-1\}$, we have
\begin{align}
u_r & = 1 + \frac{r}{d} u_{r-1} + \frac{d-r}{d} u_{r+1} & r \in \{1,2,\ldots,d-1\}\\\
u_0 & = 1 + u_1 & r = 0\\\
u_d & = 0 & r = d
\end{align}
Setting up a matrix gives us a tai-diagonal system to be solved. Instead, we go about solving this as follows.
Let $p_r = \frac{r}{d}$, $\forall r \in \{1,2,\ldots,d-1\}$. Then the equations become
\begin{align}
u_r & = 1 + p_r u_{r-1} + (1-p_r) u_{r+1} & r \in \{1,2,\ldots,d-1\}\\\
u_0 & = 1 + (1-p_0) u_1 & r = 0\\\
u_d & = 0 & r = d
\end{align}
Let $a_{r} = u_{r+1} - u_r$. Then we get
\begin{align}
p_r a_{r-1} & = 1 + (1-p_r)a_r & r \in \{1,2,\ldots,d-1\}\\\
a_0 & = -1 & r = 0
\end{align}
Note that $u_m = - \displaystyle \sum_{k=m}^{d-1} a_k$ and $u_d = 0$
\begin{align}
a_0 & = -1 & r = 0\\\
a_{r} & = \frac{p_r}{1-p_r} a_{r-1} - \frac1{1-p_r} & r \in \{1,2,\ldots,d-1\}
\end{align}
Let $l_r = \frac{p_r}{1-p_r} = \frac{r}{d-r}$
\begin{align}
a_0 & = -1 & r = 0\\\
a_{r} & = l_r a_{r-1} - (1+l_r) & r \in \{1,2,\ldots,d-1\}
\end{align}
\begin{align}
a_1 &= l_1 a_0 - (1+l_1)\\\
a_2 & = l_2 l_1 a_0 - l_2(1+l_1) - (1+l_2)\\\
a_3 & = l_3 l_2 l_1 a_0 - l_3 l_2 (1+l_1) - l_3 (1+l_2) - (1+l_3)\\\
a_m & = \left( \prod_{k=1}^{m} l_k \right) a_0 - \displaystyle \sum_{k=1}^{m} \left((1+l_k) \left( \prod_{j=k+1}^m l_j \right) \right)
\end{align}
Since $a_0 = -1$ and $l_0 = 0$, we get
\begin{align}
a_m & = - \displaystyle \sum_{k=0}^{m} \left((1+l_k) \left( \prod_{j=k+1}^m l_j \right) \right)
\end{align}
Hence,
\begin{align}
u_n & = - \displaystyle \sum_{m=n}^{d-1} a_m\\\
u_n & = \displaystyle \sum_{m=n}^{d-1} \left( \displaystyle \sum_{k=0}^{m} \left((1+l_k) \left( \prod_{j=k+1}^m l_j \right) \right) \right)\\\
u_n & = \displaystyle \sum_{m=n}^{d-1} \left( \displaystyle \sum_{k=0}^{m} \left(\frac{d}{d-k} \left( \prod_{j=k+1}^m \frac{j}{d-j} \right) \right) \right)\\\
u_n & = \displaystyle \sum_{m=n}^{d-1} \frac{\displaystyle \sum_{k=0}^{m} \binom{d}{k}}{\binom{d-1}{m}}
\end{align}
Note that
\begin{align}
u_n & = \frac{\displaystyle \sum_{k=0}^{n} \binom{d}{k}}{\binom{d-1}{n}} + u_{n+1} & \forall n \in \{0,1,2,\ldots,d-2 \}
\end{align}
The expected number of steps from one vertex away is when $n = d-1$ and hence $u_{d-1} = 2^d-1$
The expected number of steps from two vertices away is when $n = d-2$ and hence $u_{d-2} = \frac{2d(2^{d-1} - 1)}{d-1}$
The answers for the expected number of steps from a vertex and two vertices away
coincide with Douglas Zhare's comment
Initial Solution
Problems such as these fall in the category of Markov chains and one way to solve this is through first step analysis.
We shall denote the vertices of the cube by numbers from $1$ to $8$ with $1$ and $8$ being the opposite ends of the body diagonal.
Let $v_i$ denote the expected number of steps to reach the vertex numbered $8$ starting at vertex numbered $i$.
$v_1 = 1 + \frac{1}{3}(v_2 + v_4 + v_6)$;
$v_2 = 1 + \frac{1}{3}(v_1 + v_3 + v_7)$;
$v_3 = 1 + \frac{1}{3}(v_2 + v_4 + v_8)$;
$v_4 = 1 + \frac{1}{3}(v_1 + v_3 + v_5)$;
$v_5 = 1 + \frac{1}{3}(v_4 + v_6 + v_8)$;
$v_6 = 1 + \frac{1}{3}(v_1 + v_5 + v_7)$;
$v_7 = 1 + \frac{1}{3}(v_6 + v_2 + v_8)$;
$v_8 = 0$;
Note that by symmetry you have $v_2 = v_4 = v_6$ and $v_3 = v_5 = v_7$.
Hence, $v_1 = 1 + v_2$ and $v_2 = 1 + \frac{1}{3}(v_1 + 2v_3)$ and $v_3 = 1 + \frac{2}{3} v_2$.
Solving we get $$\begin{align}
v_1 & = 10\\
v_2 = v_4 = v_6 & = 9\\
v_3 = v_5 = v_7 & = 7
\end{align}$$
Hence, the expected number of steps to reach the diagonally opposite vertex is $10$.
To model this as a Markov chain, let $$S=\{(0,0,0),(0,1,0),(0,0,1),(0,1,1),(1,0,0),(1,1,0),(1,0,1),(1,1,1)\}$$ and $P$ an $8\times8$ matrix with $P_{ij}=\frac13$ if $i$ and $j$ vary in exactly one digit, $0$ otherwise. Let $\{X_n:n\geqslant 0\}$ satisfy $$\mathbb P(X_{n+1}=j\mid X_0=i_1, \ldots, X_{n-1}=i_{n-1},X_n=i)=\mathbb P(X_{n+1}\mid X_n=i)=P_{ij}. $$
Then $\{X_n:n\geqslant 0\}$ is a Markov chain, and by symmetry, both the rows and columns of $P$ sum to $1$. Since $S$ is finite, $X$ has the unique stationary distribution $\pi$ being the uniform distribution over $S$ (i.e. $\pi_i = \frac18$ for each $i\in S$). Let
$$\tau_{ij} = \inf\{n>0:X_n=j\mid X_0=i\} $$
for each $i,j\in S$. It is known that $$\mathbb E[\tau_{ii}] = \frac1{\mathbb\pi_i}, $$
so in this case the expected number of minutes to return to a vertex would be $8$.
Best Answer
This is the approach Ross Millikan gave in the comments, but with more detail.
Let the vertices of the cube be $\{0,1\}^3$, and let the ant start at $(0,0,0)$.
For each vertex $v \in \{0,1\}^3$ (aside from $(0,0,0)$), let $E(v)$ denote the expected number of edges the ant will traverse before hitting $(0,0,0)$ if it starts from $v$.
By symmetry, $E(1,0,0) = E(0,1,0) = E(0,0,1) = A$ and $E(1,1,0) = E(1,0,1) = E(0,1,1) = B$. Also, let $E(1,1,1) = C$.
If the ant is currently at $(1,0,0)$, then it will take one step to get to its next vertex, which will be $(0,0,0)$ or $(1,1,0)$ or $(1,0,1)$ with equal probability. If the ant moves to $(0,0,0)$, the ant is done. If the ant moves to $(1,1,0)$, then the expected value of the number of additional steps it needs to reach $(0,0,0)$ is $E(1,1,0)$ (by definition). Similarly, if the ant moves to $(1,0,1)$, then the expected value of the number of additional steps it needs to reach $(0,0,0)$ is $E(1,0,1)$. Hence $$E(1,0,0) = \dfrac{1}{3} \cdot 1 + \dfrac{1}{3} \cdot (E(1,1,0)+1) + \dfrac{1}{3} \cdot (E(1,0,1)+1)$$ $$A = \dfrac{2}{3}B+1$$
Using the same logic, you can work out that $$E(1,1,0) = \dfrac{1}{3} \cdot (E(1,0,0)+1)+\dfrac{1}{3} \cdot (E(0,1,0)+1) + \dfrac{1}{3}\cdot (E(1,1,1)+1)$$ $$B = \dfrac{2}{3}A+\dfrac{1}{3}C+1$$ and that $$E(1,1,1) = \dfrac{1}{3} \cdot (E(1,1,0)+1) + \dfrac{1}{3}\cdot(E(1,0,1)+1) + \dfrac{1}{3}\cdot(E(0,1,1)+1)$$ $$C = B+1$$
You can easily solve the system of equations for $A,B,C$. Finally, since the ant starts at $(0,0,0)$, its first step takes it to either $(1,0,0)$ or $(0,1,0)$ or $(0,0,1)$ with equal probability. So the answer to the question is $$\dfrac{1}{3}\cdot(E(1,0,0)+1)+\dfrac{1}{3}\cdot(E(0,1,0)+1)+\dfrac{1}{3}\cdot(E(0,0,1)+1) = A+1.$$