Defining a repeating series with a simple expression
You could just write $$ f(n) = n \pmod{4} $$ ( a loose use of notation but perfectly clear) or $$ f(n) = n - 4 \lfloor n/4 \rfloor . $$
I think the first would be kinder to your reader.
If you want something along the lines of $(-1)^n$ (which produces $1,-1,1,-1,\ldots$), first consider $$ f_1(n) = \tfrac12\left(1 - (-1)^n\right),$$ which implies that $f_1(n) = 0,1,0,1,\ldots$ for $n = 0,1,2,3,\ldots.$
Example 1
We can slow down the rate of alternation of $f_1$ by making the exponent grow slower: $$ f_2(n) = \tfrac12\left(1 - (-1)^{\lfloor n/2\rfloor}\right),$$ where $\lfloor n/2\rfloor$ is the greatest integer less than or equal to $n/2.$ This function follows the pattern $f_2(n) = 0, 0, 1, 1, 0, 0, 1, 1,\ldots$ for $n=0,1,2,3,4,5,6,7,\ldots.$
Put this together in the form $f(n) = f_1(n) + 2f_2(n)$ and you have the function you asked for. That is, let $$ f(n) = \tfrac32 - \tfrac12(-1)^n - (-1)^{\lfloor n/2\rfloor}. $$
Example 2
If you don't like the greatest-integer function (used in the expression $\lfloor n/2\rfloor$), we can work around this by fiddling with sinusoidal functions. Let $g_1(n) = \cos\left(\frac12 n\pi - \frac14\pi\right),$ so that $\newcommand{s}{\frac{\sqrt2}{2}} g_1(n) = \s,\s,-\s,-\s,\s,\s,-\s,-\s,\ldots$ for $n=0,1,2,3,4,5,6,7,\ldots.$ Then let \begin{align} f(n) &= f_1(n) + 1 - \sqrt2g_1(n) \\ &= \tfrac32 - \tfrac12(-1)^n - \sqrt2\cos\left(\tfrac12 n\pi - \tfrac14\pi\right). \end{align}
Example 3
If you're willing to use complex numbers, notice that $\cos z = \tfrac12\left(e^{iz} + e^{-iz}\right).$ Using the function notation $\exp(z) = e^z$ to make the first equation below more readable, we get \begin{align} \cos\left(\tfrac12 n\pi - \tfrac14\pi\right) &= \tfrac12 \left(\exp\left(i\left(\tfrac12 n\pi - \tfrac14\pi\right)\right) + \exp\left(-i\left(\tfrac12 n\pi - \tfrac14\pi\right)\right)\right)\\ &= \tfrac12 \left(e^{in\pi/2} e^{-i\pi/4} + e^{-in\pi/2} e^{i\pi/4}\right)\\ &= \tfrac12 e^{i\pi/4} \left(e^{in\pi/2} e^{-i\pi/2} + e^{-in\pi/2}\right). \end{align}
Then, using the facts that $e^{i\pi/2}=i,$ $e^{-i\pi/2}=-i,$ and $e^{i\pi/4}=\tfrac{\sqrt2}{2}(1+i),$ \begin{align} \cos\left(\tfrac12 n\pi + \tfrac14\pi\right) &= \tfrac12\left(\tfrac{\sqrt2}{2}(1+i)\right)\left(i^n(-i)+(-i)^n\right)\\ &= \tfrac{\sqrt2}{4}(1+i) \left((-1)^n - i\right) i^n. \end{align}
It follows that $$ f(n) = \tfrac32 - \tfrac12(-1)^n - \tfrac12(1+i)\left((-1)^n - i\right)i^n. $$
And indeed the right-hand side is also equal to $\tfrac32-\tfrac12(-1)^n-\tfrac12(1+i)(-i)^n-\tfrac12(1-i)i^n,$ an expression that was given in a comment under another answer.
Frankly, I think any of the preceding examples is far clumsier than Ethan Bolker's formula $n - 4 \lfloor n/4 \rfloor,$ which is inspired by modular arithmetic but actually uses only ordinary integer arithmetic with the greatest-integer function. That answer gets my vote.
Example 4
I have added this example after seeing that the accepted answer uses a curly bracket with multiple cases ($n$ odd and $n$ even) on the right-hand side of the equation. If two cases are OK, why not four? That is, $$ f(n) = \begin{cases} 0 \quad & n = 4k, k \in \mathbb Z \\ 1 & n = 4k+1, k \in \mathbb Z \\ 2 & n = 4k+2, k \in \mathbb Z \\ 3 & n = 4k+3, k \in \mathbb Z \\ \end{cases} $$
Just to add an alternative to the other answers: you can use generating functions to try to find an explicit solution to the recurrence relation. If $F(x)$ is the generating function for the sequence $f(n)$,
$$F(x)=\sum_{n\ge0}f(n)x^n=f(0)+f(1)x+f(2)x^2+f(3)x^3+\cdots$$
then from the recurrence given by
$$\begin{cases}f(0)=0\\f(1)=1\\f(2)=2\\f(3)=3\\f(n)=f(n-4)&\text{for }n\ge4\end{cases}$$
we have
$$\begin{align*} f(n-4)&=f(n)\\ \sum_{n\ge4}f(n-4)x^n&=\sum_{n\ge4}f(n)x^n\\ x^4\sum_{n\ge4}f(n-4)x^{n-4}&=\sum_{n\ge0}f(n)x^n-f(0)-f(1)x-f(2)x^2-f(3)x^3\\ x^4\sum_{n\ge0}f(n)x^n&=F(x)-x-2x^2-3x^3\\ x^4F(x)&=F(x)-x-2x^2-3x^3\\ (x^4-1)F(x)&=-x-2x^2-3x^3\\ F(x)&=\frac{x+2x^2+3x^3}{1-x^4} \end{align*}$$
Expanding into partial fractions, you have
$$F(x)=\frac32\frac1{1-x}-\frac12\frac1{1+x}-\frac1{1+x^2}-\frac x{1+x^2}$$
and expressing each term as a power series we arrive at
$$\begin{align*} F(x)&=\frac32\sum_{n\ge0}x^n-\frac12\sum_{n\ge0}(-x)^n-\sum_{n\ge0}(-x^2)^n-x\sum_{n\ge0}(-x^2)^n\\ F(x)&=\sum_{n\ge0}\left(\frac32-\frac12(-1)^n\right)x^n-\sum_{n\ge0}(-1)^nx^{2n}-\sum_{n\ge0}(-1)^nx^{2n+1} \end{align*}$$
each of which are valid for $|x|<1$.
Now,
$$\sum_{n\ge0}\left(\frac32-\frac12(-1)^n\right)x^n=1+2x+x^2+2x^3+\cdots$$ $$\sum_{n\ge0}(-1)^nx^{2n}=1-x^2+x^4-x^6+\cdots$$ $$\sum_{n\ge0}(-1)^nx^{2n+1}=x-x^3+x^5-x^7+\cdots$$
so it follows that
$$F(x)=\color{lightgray}{0+}\,x+2x^2+3x^3+\color{lightgray}{0x^4+}\,x^5+2x^6+3x^7+\cdots$$
Replace $n=k$ in the latter two series, then split up the first into the even and odd cases, letting $n=2k$ and $n=2k+1$, respectively, so that the generating function can be expressed as
$$F(x)=\begin{cases} \displaystyle\sum_{k\ge0}(1-(-1)^k)x^{2k}&n=2k\text{ is even}\\[1ex] \displaystyle\sum_{k\ge0}(2-(-1)^k)x^{2k+1}&n=2k+1\text{ is odd} \end{cases}$$
From this it follows that a closed form for $f(n)$ can be obtained by splitting into even/odd cases:
$$f(n)_{n\ge0}=\begin{cases}1-(-1)^{n/2}&n\text{ is even}\\2-(-1)^{(n-1)/2}&n\text{ is odd}\end{cases}$$
If I'm not mistaken, there should be a way to get an "even more explicit" form in terms of complex exponentials, but perhaps this result will suffice for your needs.