Seeing explicitly that the closed string dilaton is a scalar At level one of the closed string states have the generic form
$\gamma_{\mu\nu}\alpha_{-1}^\mu\tilde\alpha_{-1}^\nu\lvert0;p\rangle$ and are massless. It makes sense to decompose $\gamma_{\mu\nu}$ into a symmetric traceless, antisymmetric, and trace part, with the dilaton associated with the trace. That the dilaton is then a scalar seems obvious. However, we realise that the state $\eta_{\mu\nu}\alpha_{-1}^\mu\tilde\alpha_{-1}^\nu\lvert0;p\rangle$ is unphysical since the $L_{+1}$ condition requires the polarisation to be transverse, whereas here we have $\eta_{\mu\nu}p^\mu=p_\nu\neq0$. So we introduce some symmetric tensor $C_{\mu\nu}=p_\mu\overline p_\nu+\overline p_\mu p_\nu$, where $\overline p$ satisfies $\overline p\cdot p=1$ and $\overline p\cdot\overline p=0$. Then we can write down a modified state $(\eta_{\mu\nu}-C_{\mu\nu})\alpha_{-1}^\mu\tilde\alpha_{-1}^\nu\lvert0;p\rangle$ which is physical. I want to see explicitly that a state of this form is a scalar, since naively it has more than one degree of freedom. I think that this comes down to the fact all but one of the degrees of freedom are spurious, and I've shown that this state is indeed orthogonal to all physical states at levels zero and one of the closed string, except itself. Does this quality of being 'nearly spurious' confirm that we really do have a scalar? Thanks in advance.
 A: The answer to your question can be understood directly by removing $n^{\mu}$ altogether, and seeing explicitly that the only remaining quantum number is $k^{\mu}$, implying immediately that the dilaton vertex operator is a scalar.
Primarily, note that a fixed-picture dilaton vertex operator, $\hat{\mathscr{V}}$, of momentum $k^{\mu}$ that satisfies the Virasoro constraints is easy to write down (as you mention) if we choose a vector with components $n^{\mu}$ such that $k\cdot n=1$ and $n^2=0$. Up to normalisation (which is $g_c/\sqrt{D-2}$ with $D$ the number of non-compact dimensions and $g_c$ the closed string coupling, as determined by unitarity or factorisation),
$$
\boxed{\hat{\mathscr{V}}=c\tilde{c}\,\eta^{\perp}_{\mu\nu}\partial x^{\mu}\bar{\partial}x^{\nu}e^{ik\cdot x},\qquad \eta^{\perp}_{\mu\nu}=\eta_{\mu\nu}-n_{\mu}k_{\nu}-k_{\mu}n_{\nu}\,}
$$
where we have set $\alpha'=2$, and transversality, $\eta^{\perp}_{\mu\nu}k^{\mu}=0$, is manifest. (All vertex operators I write are to be considered normal-ordered in the $z,\bar{z}$ frame.)  $\hat{\mathscr{V}}$ is (as you have checked) a conformal primary of weight $(h,\tilde{h})=(0,0)$ when $k^2=0$. That is, 
$$
\tilde{L}_n\cdot \hat{\mathscr{V}}=L_n\cdot \hat{\mathscr{V}}=0,\qquad {\rm for}\qquad n\geq0,
$$
with $L_n,\tilde{L}_n$ the standard Virasoro generators of the full matter plus ghost energy-momentum tensor. (It is also true that $b_n\cdot \hat{\mathscr{V}}=\tilde{b}_n\cdot \hat{\mathscr{V}}=0$ for $n\geq0$, where $b_n,\tilde{b}_n$ are anti-ghost modes, but this will not play a dominant role here.)
Incidentally, I'm using the operator/state correspondence here, according to which:
$$
\alpha_{-1}|1\rangle\leftrightarrow i\partial x(z),\qquad \tilde{\alpha}_{-1}|1\rangle\leftrightarrow i\bar{\partial} x(\bar{z}),\qquad c_1|1\rangle\leftrightarrow c(z),\qquad \tilde{c}_1|1\rangle\leftrightarrow \tilde{c}(z)
$$
$$
|0;k\rangle\equiv e^{ik\cdot x_0}|1\rangle\leftrightarrow e^{ik\cdot x(z,\bar{z})}
$$
Your question is how to see that $\hat{\mathscr{V}}$ is a scalar. And it is a good question: clearly, a spacetime scalar (from the spacetime viewpoint) should only depend on $k^{\mu}$ and nothing else (and different scalars are distinguished by their mass, and there may in general be a degeneracy for fixed mass, compactification data may also enter, etc., which will also enter in the expression for the mass and the string coupling, $g_c$, but this is irrelevant for the dilaton). So why is $\hat{\mathscr{V}}$ a scalar, given we introduced a vector $n^{\mu}$ which seemingly might introduce new parameters into the state (it doesn't) while breaking manifest spacetime covariance (which it does as it singles out a distinguished direction in spacetime)? We can answer this in a simple way by restoring manifest covariance (i.e. removing the apparent $n^{\mu}$-dependence), as follows.
Recall that the physical state conditions for any vertex operator $\hat{\mathscr{V}}$ are that it lives in the cohomology of the BRST charge, $Q_B$. The crucial point is that not all states in the cohomology class of the dilaton are conformal primaries, i.e. not all of them are invariant under conformal transformations, but are perfectly acceptable physical states nevertheless as they are annihilated by $Q_B$ (and cannot be written as $Q_B\cdot \mathscr{W}$ for some $\mathscr{W}$). To bring this point into plain view, note that we can hop from one state in the dilaton cohomology class to another (i.e. from one description of the same physical state to another) by adding BRST exact terms, 
$$
\hat{\mathscr{V}}\rightarrow \hat{\mathscr{V}}'=\hat{\mathscr{V}}+Q_B\cdot \mathscr{W},
$$ 
so if the dilaton is a scalar there must be a choice of $\mathscr{W}$ for which the $n^{\mu}$ dependence drops out, leaving only the dependence on $k^{\mu}$. Of course, if $\mathscr{V}$is physical so will $\hat{\mathscr{V}}'$ be physical. The choice that does the job is to take: 
$$
\mathscr{W}=\big(c\,n_{\mu}\partial x^{\mu}-\tilde{c}\,n_{\mu}\bar{\partial} x^{\mu}\big)e^{ik\cdot x}.
$$ 
Computing $Q_B\cdot \mathscr{W}$ (which I leave as an exercise, see Polchinksi's book for the explicit expression for $Q_B$) leads to the new vertex operator, 
\begin{equation}\label{eq:V'=V+exact}
\begin{aligned}
\boxed{\hat{\mathscr{V}}'=\Big(c\tilde{c}\,\eta_{\mu\nu}\partial x^{\mu}\bar{\partial}x^{\nu}-\tfrac{1}{2}c\partial^2c+\tfrac{1}{2}\tilde{c}\bar{\partial}^2\tilde{c}\Big)e^{ik\cdot x}\,}
\end{aligned}
\end{equation}
Clearly, from the above $\hat{\mathscr{V}}'$ and $\hat{\mathscr{V}}$ describe the same physical state. Covariance is restored (we have completely removed the ugly dependence on $n^{\mu}$) and most importantly the state is manifestly completely specified by the choice of momentum: $k^{\mu}$. So this is a scalar, given that there is no other quantum number characterising the state. 
Your question is therefore answered, but I should mention a drawback. Although covariance is restored, the vertex operator $\hat{\mathscr{V}}'$ is not a conformal primary (it is not annihilated by $L_n,\tilde{L}_n$ for $n>0$), and so is not conformally invariant. This is clearly no problem as BRST invariance is a more general concept; we don't need conformal invariance, although it does help to have it for many computations. 
Finally, I will mention the analogue of $\hat{\mathscr{V}}$ and $\hat{\mathscr{V}}'$ in the integrated picture, that I denote by $\mathscr{V}$ and $\mathscr{V}'$ respectively. Because this involves some additional technology that would take a few more paragraphs to explain (given it doesn't exist in standard textbooks), and because the answer to your question has already emerged, I will simply state the results*:
$$
\boxed{\mathscr{V} = \int d^2z \,\eta^{\perp}_{\mu\nu}\partial x^{\mu}\bar{\partial}x^{\nu}e^{ik\cdot x}\,}
$$
for the conformal primary, and:
$$
\boxed{\mathscr{V}' = \int d^2z\,\Big(\eta_{\mu\nu}\partial x^{\mu}\bar{\partial}x^{\nu}-R_{z\bar{z}}\Big)e^{ik\cdot x}\,}
$$
for the covariant (non-primary) integrated picture vertex operator. Again, you see that $\eta_{\mu\nu}$ appears and not $\eta_{\mu\nu}^{\perp}$ in $\mathscr{V}'$, so the $n^{\mu}$ dependence again drops out.  Here $R_{z\bar{z}}$ is the worldsheet Ricci tensor. Care is needed in interpreting these expressions, and the "coordinates" $z,\bar{z}$ are not really worldsheet coordinates but are best interpreted as frame coordinates, see Polchinski for all the glory details where everything I have covered is explained in much greater detail. I will leave it at this.
*(This elaborates on @Nogueira's comment about it not being possible to write down covariant physical states without ghosts -- the point I'm highlighting here is that in the integrated picture one can instead write down covariant vertex operators without ghost BRST technology.)
