Am I attracting Pluto? My question is simple: as the title says, do I exert a gravitational force on distant objects, for example, Pluto? Although it is a very small force, it is there, right?
This leads me to the question, am I exerting a gravitational force on everything in the Universe, for example the farthest galaxy that we know of?
 A: The mass that makes up your body and everything inside your body is attracting Pluto. If you did not exist, then that mass would take other forms, and it would still attract Pluto by the same amount. I don't want to offend, but you know where all your mass came from, and where it's going. It would likely be in one of those forms had you not been conceived, and would still attract Pluto. Pluto is not affected by the fact that you are alive in any way.
Let us assume that you have a mass of $60\ \mathrm{kg}$. Pluto is $7.5 \times 10^9\ \mathrm{km}$ away, that is 4.5 light-hours. Pluto's mass is $1.3 \times 10^{22}\ \mathrm{kg}$.
The force that you attract Pluto by (and Pluto attracts you by) is:
\begin{align}
    F &= G\frac{M_1M_2}{d^2} \\
      &= \left(6.7 \times 10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}\right)\frac{60\ \mathrm{kg}\left(1.3 \times 10^{22}\ \mathrm{kg}\right)}{\left(7.5 \times 10^{12}\ \mathrm m\right)^2} \\
      &= 9.2 \times 10^{-13}\ \mathrm N
\end{align}
For comparison, this is the force that approximately 100 bacteria were to exert on your hand if you were to hold them.
Though other posters mention your light-cone, I doubt that your mass differs significantly over the course of a 4.5 hour period. Technically, the force that you exert on Pluto right now is your mass from 4.5 hours ago. Don't eat or go to the toilet for 4.5 hours, and you'll be fine.
A: Antonio --
sure: your mass affects Pluto.
Here's a line of thought that might get your head around this amazing concept:
You probably agree that the Earth as a whole affects Pluto .. right?
Simply, consider the earth chopped up in to little pieces each about the size and mass of yourself .. say 100kg.
(Funnily enough .. there's actually NOT THAT MANY of them.   If you label them on three axis, there's only roughly 50 million across on each axis.)
OK - so you're picturing the Earth as very many "Antonio-like pieces". Indeed there's about 10e23 ALPs (ALP == Antonio-like piece!).
Now, you know and agree that all those together .. affects Pluto. Right?
So, if you get in to this way of looking t things, it's probably easy to see that - of course - you (the particular ALP that is you!) affects Pluto.
After all, it would be ridiculous to say that your ALP does "not" affect Pluto but somehow all the rest do. Right?
So, this is just a good way to look at such "really surprising" factoids, as that your particular ALP affects Pluto.
Now here's a really interesting further thought.  For me, it's actually totally amazing how STRONG, yes, I said STRONG, merely one ALP is.
Consider how frickin' big Pluto is, and how fast moving it is.  Incredibly, it only takes the collection of ALPs we have here on Earth to actually affect Pluto.  that's... astounding.
Don't even mention the Sun. The Sun only has about 10e28 ALPs (hope you're not getting hot there!)
But -- absolutely incredibly -- the Sun can "throw around" PLUTO , for goodness sake (Pluto is A WHOLE PLANET!!!!) .... the Sun can fling Pluto around in a circle at astounding speed.
Imagine just sort of theoretically you had a wire or something that could FLING A PLANET AROUND.  I mean it's nuts right?
But you ONLY NEED about 10e28 of yourself, to, fling around Pluto - and at that incredible distance.
Yes, you are bouncing Pluto around right now.  It's nuts!
A: Let's make this question a bit more operationally meaningful by asking if you can change the state of Pluto by choosing to do something here. As mentioned in the other answers, Pluto would feel the same force due to your mass even if you didn't exist, because the matter you consist of would be present on Earth anyway. However, you can still chose to move in a certain way and one can then consider the effect such a choice has on Pluto.
If you move then the distance between you an Pluto changes, if the distance changes from $d$ to $d+u$, the force will change. If $F(r)$ is the magnitude of the force exerted by you on Pluto, then we have:
$$\begin{split}
F(d+u) - F(d) &= G M_{\text{Pluto}}M_{\text{Antonio}}\left[\frac{1}{(d+u)^2}-\frac{1}{d^2}\right]\\&\approx  G M_{\text{Pluto}}M_{\text{Antonio}}\left(-2\frac{u}{d^3}+3\frac{u^2}{d^4}\right)
\end{split}
$$
As pointed out in the comments by Dan and SchighSchagh, we must also take into account that the Earth moves in the opposite direction (actually, it's only part of the Earth as it cannot be treated as a rigid object, but that doesn't matter here), the center of mass doesn't change while the change in the force exerted on Pluto due to all the changes caused by the jump is to first order in $u$ due to the change in the center of mass. So, as pointed out by SchighSchagh, there is no net first order contribution.
The leading effect on Pluto is thus due to the second order term. The contribution due to the recoil of the Earth can then be ignored because the displacement squared times the mass for the Earth is now suppressed by the mass ratio of Antonio and the Earth relative to Antonio's contribution. We thus have:
$$F(d+u) - F(d) \approx 3G M_{\text{Pluto}}M_{\text{Antonio}}\frac{u^2}{d^4}
$$
To be precise we need to take into account that the change in the force experienced by Pluto now is due to the value of $u$ about 4.5 hours ago, so we need to use the so-called "retarded" value of the variable. Suppose then that Pluto will be overhead in 4.5 hours from now and you jump up to a height of $h$. The variable $u$ as a function of time will then be given by:
$$u(t) = -\sqrt{2 g h} t + \frac{1}{2}g t^2$$
for $0\leq t \leq \frac{2\sqrt{2 g h}}{g}$ 
The component of Pluto's momentum in the direction away from Earth will thus increase due to the jump by an amount of:
$$\Delta P_{\text{Pluto}}=\frac{3G M_{\text{Pluto}}M_{\text{Antonio}}}{d^4}\int_{0}^{\frac{2\sqrt{2 g h}}{g}}\left(\sqrt{2 g h} t - \frac{1}{2}g t^2\right)^2dt=\frac{4G M_{\text{Pluto}}M_{\text{Antonio}}}{15\sqrt{g}d^4}(2h)^{\frac{5}{2}}$$ 
Putting in the numbers here yields:
$$\Delta P_{\text{Pluto}}=7.9\times10^{-39}\frac{M_{\text{Antonio}}}{60\text{ kg}}\left(\frac{h}{\text{meter}}\right)^{\frac{5}{2}}\text{ Ns}$$
So, it seems that there is very small but real physical effect on Pluto. However when an  extremely small amount of momentum is transferred, the physical state of the system may actually not change at all. This is due to the fact that the momentum of a system doesn't have a precise value due to quantum mechanics. A good example where you can see this effect at work is in certain variants of the double slit experiment where photons going through the slits will be bounced off mirrors before they hit the screen. If the imparted momentum on the mirrors would change the physical state of the mirror or the rest of the universe, then the interference pattern would vanish, because the information about which path the photon took will in principle exist. But in such experiments the interference pattern remains visible, which is experimental proof that the physical state of the rest of the universe actually does not change despite the momentum transfer.
To see if this effect is relevant, one has to give an approximate quantum mechanical description of Pluto's center of mass motion. Obviously, if Pluto were in some momentum eigenstate then the small momentum change would cause it's physical state to change, but obviously Pluto is not in such a state. A good approximation is obtained as follows. Pluto is not an isolated object, it receives energy from the Sun, its surface is at approximately 30 K. So, we can model it by assuming that all the degrees of freedom of Pluto are in a thermal bath at 30 K, and one of these degrees of freedom is its center of mass. What then happens is that due to the interactions with the thermal bath, the quantum mechanical uncertainty of the momentum of the center of mass is limited to about:
$$\Delta P_\text{QM}\approx \sqrt{M_{\text{Pluto}}k T}= 2.3 \text{ Ns}$$
So, the center of mass can be pictured as being described by an unknown wavefunction that in momentum space has a typical width of a few newton seconds. Since this is much larger than the  transferred momentum, the state it would be in had you not jumped and the state it is in due to the jump have an overlap that is almost identical to 1. For the state to have changed unambiguously, the overlap between the two states should be zero. In terms of probabilities, one can say that Pluto will fail to detect whether or not you jumped with a probability of almost 1.
A: While the atoms that make up your body are exerting gravitational force on very distant objects you as an entity are only exerting gravitational force on objects out to about 14 light years distance (assuming the age shown in your profile is correct). 
Because the "speed of gravity" is the speed of light. And toward the outer edge of that sphere the forces are controlled by your birth mass rather than your current mass.
A: You -- and more to the point the matter you are made of -- are part of the Earth. The 65% water content of your body was part of the oceans before you were born and will be again. Likewise the other 35% of minerals and other substances come from Mother Earth. Thus whatever gravitic influence you have is both imperceptible and inconsequential in the context of the gravitic influence your birth planet.
Now if you were to take your mass and leave Earth and "stand" somewhere else in the solar system (other than Pluto) then you probably would have some influence but again it would likely be imperceptible to any measurements we can produce.  

UPDATE: The recent discovery of gravitational waves reminded me of this question. After mulling my answer over again for a bit I came to realize I needed to add a small addendum... 
While nothing in my original answer has changed, the reader should also factor in that the Earth is constantly losing mass at the rate of ~0.000000000000001 (~1.0E-15) % per year. So on an annual basis the Earth loses a femto-scale degree of pull on Pluto, and with it the person who asked the question gains an infinitesimal fraction of pull relative to the remaining mass of the Earth. 
Additionally the body of Pluto is going through the same kinds of changes that Earth is, except most of its atmosphere is frozen so it probably is gaining a net mass from added cosmic dust. Thus in the far future Pluto will likely pull the Earth more than the other way around. By then the original questioner's body will be ancient timeworn dust (still integral to the mass of the Earth) and the answer will still basically remain the same.
Kind of gives a whole new perspective to the lyric "All we are is dust in the wind." 
A: The simple answer to both questions is, yes, you exert a force on Pluto and on the universe.  By simple, I mean that the Pluto question is treated as a "two body" problem (and nothing else). This answer is obtained using the formula given by dotancohen.
For the universe question, it should be easy to see that the same process can be applied to you and any other "body" in the universe.  Therefore, you exert a force on every thing in the universe in proportion to the product of your and the other masses and inversely as the square of the distance between you and the other masses.       
