The relativistic effects are very small, but not that small, let's compute them explicitly, it's a good exercise.
Basically the first thing you would think of would be to just compose the gravitational time dilation with the special-relativistic "doppler" time dilation; what you would get in this particular case would not be wrong, but it's not correct to assume a priori that such a thing makes sense. Therefore let's just do the whole calculation and then check that it's ok.
The metric for a weak static field such as that of the sun is given by \(g_{00} = - (1 + 2 \Phi)\) where \(\Phi\) is the gravitational potential and all other components are equal to the flat space ones (\(g_{\mu\nu} = \eta_{\mu\nu}\) for all other indices). Placing Mercury at position (R,0,0) and velocity (meaning \(dx^i/dt\) with t coordinate time) (0,V,0) then let's write down \(\frac{dx^\mu}{dt}\) (which is not a vector, because t is not a scalar):
\[ (1,0,V,0) \]
pretty self-explanatory. Then the four velocity \(u^\mu\) must be proportional to this thing, but also be normalized as \(u^\mu u_\mu = - 1\). Then taking the square of the above "vector" (using the metric), we get \(- (1+2\Phi) + v^2\), meaning our 4-velocity is actually:
\[ u^\mu = (1 + 2\Phi - v^2)^{-1/2} (1,0,V,0) \]
that's very useful, because we can read the time dilation factor as the 0 component of \(u\):
\[ u^0 = \frac{dt}{d\tau} = \frac{1}{\sqrt {1+2\Phi-v^2 }} \]
notice the similarity with the usual Doppler time dilation gamma factor; the only addition is the gravitational potential. Since we are talking about very small stuff, we can Taylor-expand:
\[ \sim 1 - \Phi + \frac{V^2}{2} \rightarrow 1 + \frac{1}{c^2} (-\Phi + V^2/2) \]
Which is exactly what we would get by composing / summing gravitational and SR time dilation - so we see that this operation is only lecit in the limit of small \(\Phi\) and \(V\). I've reintroduced factors of \(c\).
One could now substitute the orbital radius and velocity of Mercury (assuming a circular orbit, which is wrong for Mercury but w/e) and get the answer but there's an important simplification: Mercury is in orbit! Therefore:
\[ \frac{1}{2} m V^2 = \frac{G m M}{2R} \]
that's actually equivalent to Newton's law as you can readily check and takes the name of virial theorem. Helps a lot because our time dilation factor becomes:
\[ u^0 = 1 + \frac{1}{c^2} \left( \frac{GM}{R} + \frac{V^2}{2} \right) = 1 + \frac{3}{2} \frac{1}{c^2} \frac{GM}{R} \]
So our time dilation factor ends up being
\[ 1 + 5 \cdot 10^{-8} \]
meaning that over a Mercurian year of 87 days an observer on Mercury would measure around 0.3 seconds less of proper time.