As far as you don't require Lorentz invariance, it is perfectly OK to work in four dimensions.
How many states do you have at each level ? You can convince yourself, as an exercise, that at level $k$, the number of states is the coefficient of $x^k$ in the function $$\prod\limits_{m \geq 1} \frac{1}{(1-x^m)^2} = 1 + 2x + 5x^2 + 10x^3 + 20 x^4 + \dots $$
Now clearly those numbers do not coincide with the dimensions of the irreducible representations of $SO(3)$, which are the odd integers. The reason is simply that the representations to which the string states belong are not irreducible. The 10-dimensional representation is the sum of a spin 3 and a spin 1.
One way to see that is to refine the generating function given above. For that, include an additional variable $s$ that will "count the spin": $$\prod\limits_{m \geq 1} \frac{1}{(1-s^2x^m)(1-s^{-2}x^m)} = 1+\left(s^2+\frac{1}{s^2}\right)
x+\left(s^4+\frac{1}{s^4}+s^2+\frac{1}{s^2}+1\right)
x^2+\left(s^6+\frac{1}{s^6}+s^4+\frac{1}{s^4}+2
s^2+\frac{2}{s^2}+2\right)
x^3+\left(s^8+\frac{1}{s^8}+s^6+\frac{1}{s^6}+3
s^4+\frac{3}{s^4}+3 s^2+\frac{3}{s^2}+4\right)
x^4 \dots $$
The coefficient of $x^3$ is the sum of the character of a spin 3, $s^6+\frac{1}{s^6}+s^4+\frac{1}{s^4}+2
s^2+\frac{1}{s^2}+1$, and a spin 1, $s^2+\frac{1}{s^2}+1$.
For another example, let us consider level 4. Here the character is $$s^8+\frac{1}{s^8}+s^6+\frac{1}{s^6}+3
s^4+\frac{3}{s^4}+3 s^2+\frac{3}{s^2}+4$$. To decompose it, we can start with the highest weight, $s^8$. This is spin 4. We remove the spin 4 character, $s^8+\frac{1}{s^8}+s^6+\frac{1}{s^6}+
s^4+\frac{1}{s^4}+1 s^2+\frac{1}{s^2}+1$ and we are left with $2
s^4+\frac{2}{s^4}+2 s^2+\frac{2}{s^2}+3$. We see that the highest weight is $2s^4$, so we have two spin 2. Removing $2 \left(
s^4+\frac{1}{s^4}+ s^2+\frac{1}{s^2}+1 \right)$, we are left with $1$, which is a spin 0. Conclusion: the level 4 is one spin 4, two spin 2 and one spin 0. The dimensions add up to $9 + 2 \cdot 5 + 1 = 20$.
You can play the same game at any level, and observe that except at level 1, you can always cast the spectrum into (in general reducible) representations of $SO(3)$.
Edit : how to tell which states give which spins
For that, you can refine even more the generating function. Let us take a different fugacity for each oscillator level,
$$\prod\limits_{m \geq 1} \frac{1}{(1-s^2x^m \alpha_{-m})(1-s^{-2}x^m \alpha_{-m})} =1+x \left(\alpha _{-1} s^2+\frac{\alpha _{-1}}{s^2}\right)+x^2
\left(\alpha _{-1}^2+\alpha _{-1}^2 s^4+\frac{\alpha
_{-1}^2}{s^4}+\alpha _{-2} s^2+\frac{\alpha
_{-2}}{s^2}\right)+x^3 \left(2 \alpha _{-2} \alpha _{-1}+\alpha
_{-1}^3 s^6+\frac{\alpha _{-1}^3}{s^6}+\alpha _{-2} \alpha _{-1}
s^4+\frac{\alpha _{-2} \alpha _{-1}}{s^4}+\alpha _{-1}^3
s^2+\alpha _{-3} s^2+\frac{\alpha _{-1}^3}{s^2}+\frac{\alpha
_{-3}}{s^2}\right)+ \dots$$
Let us look for instance at level 3. You see that the spin 3 is made by states of the form $\alpha_{-1}^3$ and $\alpha _{-2} \alpha _{-1}$, while the spin 1 is made of states of the form $\alpha_{-3}$ and $\alpha _{-2} \alpha _{-1}$.
