Let $w_n$ be the number of random walks of length $n$ in a $d$-ary tree that start and end at the root, and let $s_n$ be the number of such walks that do not visit the root at any point during the middle. Then $w_n$ and $s_n$ satisfy the relations $ w_n \;=\; \sum_{i_1+\cdots+i_k = n}s_{i_1}\cdots s_{i_k}\qquad\text{and}\qquad s_n = w_{n-2}d. $ The first is from the fact that any walk of type $w$ is a concatenation of finitely many walks of type $s$. The second is from the fact that a walk of type $s$ of length $n$ can be viewed as a walk of length $n-2$ of type $w$ based at one of the $d$ child vertices of the root.
It is easy to check that $ w_{2n} \;=\; C_n d^n \qquad\text{and}\qquad s_{2n} \;=\; C_{n-1}d^n $ are solutions to these recurrence relations, where $C_n$ is the $n$th Catalan number.
In particular: $ w_2 \;=\; d,\quad w_4 = 2d^2,\quad w_6 = 5d^3,\quad w_8 = 14d^8,\quad \ldots $
Edit: The above formulas work for the infinite rooted $d$-ary tree, but it seems that you were asking about a rooted tree in which the root has $d$ children, and every other node has $d-1$ children.
In this case, the formulas above work for random walks in each of the $d$ basic subtrees, with $d-1$ in place of $d$. For random walks based at the root, it follows that $ s_{2n} \;=\; C_{n-1}(d-1)^{n-1}d $ and $ w_{2n} \;=\; \sum_{i_1 + \cdots + i_k = 2n} s_{i_1}\cdots s_{i_k} $ I'm not sure there's a simpler formula for $w_{2n}$ than this.