First of all, there are several typos in your calculations (e.g. it should read $\int_0^t W_s^2 \,ds$ instead of $\int_0^t W_t^2 \, ds$). Your calculation goes wrong when you write
$$\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$
(I don't get what you did in this last step - you want to calculate $\mathbb{E}(W_t^4)$; so why replace it with $3t^4$?)
Note that applying Itô's lemma is overkill: Since $W(_t)_{t \geq 0}$ is a Wiener process, we know that $W_t \sim N(0,t)$ (i.e. $W_t$ is Gaussian with mean $0$ and variance $t$) and the moments of Gaussian random variables can be calculated explicitly. However, if you really want to invoke Itô's formula, then it goes like that: By Itô's formula, we have
$$W_t^4 = 4 \int_0^t W_s^3 \, dW_s + 6 \int_0^t W_s^2 \, ds. \tag{1}$$
Since $(W_s^3)_{s \geq 0}$ is properly integrable, we know that the stochastic integral
$$M_t := \int_0^t W_s^3 \, dW_s$$
is a martingale and therefore $\mathbb{E}M_t = \mathbb{E}M_0=0$. Taking expectation in $(1)$ yields
$$\mathbb{E}(W_t^4) = 6 \int_0^t \mathbb{E}(W_s^2) \, ds$$
by Fubini's theorem. Finally, since $\mathbb{E}(W_s^2)=s$, we get $\mathbb{E}(W_t^4) = 3t^2$.
It's enough that $f$ be Lebesgue measurable and $\int_0^t [f(s)]^2\,ds<\infty$ for each $t$. Indeed, in this case your stochastic integral is well defined. Moreover, if we set $A_t:=\int_0^t[f(s)]^2\,ds$, then $Y$ is a (mean zero) continuous martingale with quadratic variation $\langle Y\rangle_t=A_t$.
Let $\tau(t):=\inf\{s:A_s>t\}$. Then $Z_t:=Y_{\tau(t)}$ is a continuous martingale with quadratic variation $t$; by Levy's theorem, $Z$ is Brownian motion. Because $A$ is deterministic, $Y_t=Z_{A_t}$, $t\ge 0$, is a deterministic re-parametrization of a Gaussian process, and is therefore also a Gaussian process.
Best Answer
As noted in the comments, this is a random variable, but it looks like you're looking for an equality similar to something like $$\frac{1}{2}(W_T^2 - T) = \int_0^T W_s \,dW_s.$$
In other words, you're looking for a smooth function $f:\mathbb{R}^2 \to \mathbb{R}$ so that $$f(W_T,T) = \int_0^T s^2 W_s\,dW_s.$$ This interpretation of your question is mentioned in the comments, and I think that this is the sort of thing you're looking for. By Itô's Lemma, we would need $f$ to satisfy $$\begin{cases}\displaystyle\frac{\partial f}{\partial x} = t^2 x, \\ \displaystyle\frac{\partial f}{\partial t} + \frac{1}{2} \frac{\partial^2 f}{\partial x^2} = 0, \\ f(0,0)=0 \end{cases}$$ which you can check has no solution since the first equation implies $\frac{\partial^2 f}{\partial x^2}$ is constant with respect to $x$, thereby making the second equation impossible.
To answer the question about "indefinite integration for Itô integrals," the answer is that you can "find an antiderivative" (i.e. an answer of the form $f(W_T,T)$) when you're integrating $\frac{\partial f}{\partial x}$ where $f$ satisfies the diffusion equation $$\frac{\partial f}{\partial t} = -\frac{1}{2}\frac{\partial^2 f}{\partial x^2}.$$ This can be seen from Itô's Lemma.