Finding minimal and complete test sets for circuits
Yes.  One can construct a Majella sequence of length $n \cdot 2^n + 1$, recursively.  It's easy to see you can't do any better than this, so this is optimal.

Definition. I'll say that a sequence $x_0,x_1,\dots,x_m$ of $n$-bit values is a $n$-bit Majella sequence if $x_0=x_m$, and for each $y \in \{0,1\}^n$ and each value $e \in \{0,1\}^n$ with a single bit set, the pair $y,y\oplus e$ appears (consecutively) somewhere in the sequence.

Recursive construction. Let $w_0,w_1,\dots,w_m$ be a $n-1$-bit Majella sequence (for $n-1$-bit values).  Consider the following sequence:

$$0w_0, 0w_1,\dots, 0w_{m-1}, 0w_m, 1w_m, 1w_{m-1}, \dots, 1w_1, 1w_0, 0w_0,$$

where $0w$ represents the concatenation of $0$ and $w$.  This sequence is not yet a Majella sequence, but we can modify it to be one.  In particular, for each value $w \in \{0,1\}^{n-1} \setminus \{w_0\}$, replace the first occurrence of $0w$ in the sequence with $0w,1w,0w$ (increasing the length of the sequence by 2).

After this modification, the result is a $n$-bit Majella sequence.

This construction is inspired by the standard recursive construction of a Gray code, but tweaked slightly to satisfy your particular requirements.

Length of the recursive construction.  Let $L(n)$ be the length of the resulting sequence (i.e., the value of $m+1$).  It satisfies the recurrence relation

$$L(n) = 2L(n-1) + 2^n - 1$$

with $L(1) = 3$.  This solves to

$$L(n) = n \cdot 2^n + 1.$$

Algorithms. I don't know if there are any efficient, low-space algorithms for enumerating through such a sequence, like there are for Gray codes.