Solution:

Let the three outfits be $A$, $B$, and $C$. For each performance, the performer can choose any of the three outfits, so there are $3^7$ total ways to assign outfits to the seven performances.

However, we require that each outfit is worn at least once. We use the principle of Inclusion-Exclusion.

Let $S$ be the set of all assignments. For each outfit $X$ ($A$, $B$, or $C$), let $S_X$ be the set of assignments where outfit $X$ is not worn at all.

The number of assignments where at least one outfit is not worn is:

$|S_A| + |S_B| + |S_C| - |S_A \cap S_B| - |S_A \cap S_C| - |S_B \cap S_C| + |S_A \cap S_B \cap S_C|$

Now,
- $|S_X|$: If outfit $X$ is not worn, only two outfits are available for all performances, so $2^7$ ways.
- $|S_X \cap S_Y|$: If two outfits are not worn, only one outfit is available, so $1^7 = 1$ way.
- $|S_A \cap S_B \cap S_C|$: No outfit is worn, which is impossible, so $0$ ways.

There are 3 choices for $X$ and 3 choices for pairs $X, Y$.

So, by Inclusion-Exclusion:

Number of ways $= 3^7 - 3 \times 2^7 + 3 \times 1$

Calculate:
- $3^7 = 2187$
- $2^7 = 128$
- $3 \times 128 = 384$
- $3 \times 1 = 3$

So,

Number of ways $= 2187 - 384 + 3 = 1806$

Final Answer: $1806$