$$p_{00}^{2n}={2n\choose n}p^n(1-p)^n= {1\over 4^n}{2n\choose n}\times [4p(1-p)]^n\leq [4p(1-p)]^n.$$
The sum of the right hand side is a convergent geometric series if $p\neq 1/2$.
Added: I think I understand your problem better now.
I hope this is what you want; let me know if anything is unclear.
You want to know how $X_n\to\infty$ implies $\sum_n p^n_{00}<\infty$.
To make a direct connection between the sum and the random walk, let
$N=\sum_{n=0}^\infty 1_{(X_n=0)}$ denote the total number of visits to state $0$.
Then $\sum_n p^n_{00}=E(N)\leq\infty$.
Now define the return time to zero as $T:=\inf(n>0: X_n=0)$.
By the strong Markov property, the random variable $N$ is geometric with probability of success $P(T=\infty)$.
If $P(T=\infty)=0$, then $P(N=\infty)=1$ which contradicts $X_n\to\infty$.
Therefore, we have $P(T=\infty)>0$ and $E(N)={1\over P(T=\infty)}<\infty$, which shows that the random walk is transient.
Further calculations would give the explicit formula $E(N)=1/(1-2p)$.
The vertices of a cube(oid) can be labelled with 3-digit binary numbers since there are $8$ corners and $2^3=8$. To keep things simple lets say that we are considering the unit cube and the binary number correspond directly to the coordinates, so that vertce $000$ is at $(0,0,0)$, vertex $010$ is at $(0,1,0)$ and so forth. Taking one step on the random walk is equivalent to picking one digit with uniform probability of $1/3$ from our binary number and flipping it.
W.l.o.g. let us consider starting at $000$, let us say that the expected time to return to $000$ is $x$. Now consider the three vertices $001$,$010$ and $100$, they are all one step away from the start. Let's say the expected time to return to the start from here is $y$. Similarly consider the vertices with two ones $110$, $101$ and $011$. These are all two steps away from the start. Let us say that the expected time to get back to $000$ from here is $z$.
How can we link all of these together? Well, if we are at one of the vertices $001$, $010$ or $100$ we have a $1/3$ probability of returning to the start and a $2/3$ probability to go to one that is two steps away. This gives us
$$y = \frac{1}{3} + \frac{2}{3}(z+2)$$
To make sense of this, see that we have a 1/3 probability to be finished, otherwise we are at a vertex two steps away (whose expected time is $z$) but we took two steps to get there (hence $z+2$). We can perform a similar process for the vertices $011$, $101$ and $110$. Here we have a $2/3$ probability to return to a vertex one step from the start and a $1/3$ prob. to go to $111$, but at $111$ our only option is to back to one of the vertices two steps away and so the corresponding equation reads
$$z = \frac{2}{3}(y+1) + \frac{1}{3}(z+2)$$
Using these we can solve to get $y=7$. Now clearly from $000$ we can only reach vertices one step away, for which we already know the expected time to return to $000$ is 7. Thus our expected time to return to $000$ is $x=y+1=8$.
Although our discussion focused on starting from $000$ there was nothing special about this choice other than to keep the number of ones equal to the steps away from the start and thus it holds for any starting vertex the expected number of steps to return is $8$.
I decided to check whether this answer is correct using Monte-Carlo. With 1,000,000 trials the answer came out as 7.99 (3 s.f.).
Best Answer
Let $\mathsf{P}_v$ be the law of the random walk $X_n$ starting at $v$ (i.e., $X_0=v$) and let $T_a:=\inf\{n\ge 1:X_n=a\}$. We will use the fact that for $a<0<b$, $$ \mathsf{P}_0(T_a<T_b)=\frac{b}{b-a}. $$
It is clear that $\mathsf{P}_0(M_1\ge 1)=1/2$. For $k>1$, \begin{align} \mathsf{P}_0(M_k\ge 1)&=\frac{1}{2}\mathsf{P}_0(M_k\ge 1\mid X_1=1)+\frac{1}{2}\mathsf{P}_0(M_k\ge 1\mid X_1=-1) \\ &=\frac{1}{2}\mathsf{P}_1(M_k\ge 1)+0=\frac{1}{2}\mathsf{P}_0(T_{k-1}<T_{-1})=\frac{1}{2k}. \end{align}
Using the strong Markov property, \begin{align} \mathsf{P}_0(M_k\ge 2)&=\mathsf{P}_0(M_k\ge 2,M_k\ge 1)=\mathsf{E}_0[\mathsf{P}_k(M_k\ge 1)1\{M_k\ge 1\}] \\ &=(1-\mathsf{P}_0(M_k\ge 1))\mathsf{P}_0(M_k\ge 1). \end{align}
An interesting implication of this result is that $$ \mathsf{E}_0 M_k=\sum_{n\ge 1}^{\infty}\mathsf{P}_0(M_k\ge n)=1. $$