Let's introduce the function $g(x) = f(x) + 3 = a\tan x - bx^5 + cx$. According to the problem statement $f(-3)=7$, we have $g(-3)=f(-3)+3=10$.

Firstly, let's consider the transformation of $g(x)$ when $x$ is replaced by $-x$:
\begin{align*}
g(-x) &= a\tan (-x) - b(-x)^5 + c(-x)\\
&= -a\tan x + bx^5 - cx\\
&= -(a\tan x - bx^5 + cx)\\
&= -g(x).
\end{align*}

From the last equation, we can see that $g(x)$ is an odd function since $g(-x) = -g(x)$ holds for all $x$ in the domain.

Now, knowing that $g(-3) = 10$, and using the fact that $g(x)$ is odd, we can find the value of $g(3)$:
\begin{align*}
g(3) &= -g(-3)\\
&= -10.
\end{align*}

Hence, we can determine the value of $f(3)$ using $g(3)$:
\begin{align*}
f(3) &= g(3) - 3\\
&= -10 - 3\\
&= -13.
\end{align*}

Therefore, the value of $f(3)$ is \boxed{-13}.